Chapter 21 (Reading from Young&Friedman): Charge, force, & electric field. Last update: 2/23/09

21-1: a) The "coulomb" is the amount of charge that would generate an electric force of 9x109 N when two equal 1C charges are 1 meter apart. In other words the "coulomb" is defined to make k=9x109 N.m2 /C2

b) 6.25 x1018 electrons in 1 C.c) This problem requires some intelligent guessing. Ping pong balls are mostly paper which is mostly carbon (Atomic mass=12 g/mole). A 2-g ball will contain 1/6 of a mole. Since carbon atoms have 6 electrons, that means there is about 1 mole (6.02 x 1023) electrons in the ping-pong ball. From here it is relatively easy to determine that 1C is about 10-4% of the electrons in the ping-pong ball. It would require, of course, an unreasonably large amount of force to remove these many electrons from any object.d) 3 x 109 esu = 1C. The technique is to compare "k" in MKS units to "k" in cgs units.e) proton: u-u-d and neutron: u-d-d

21-2: These were discussed in class. Check your notes.

21-3: a) 9.2 x 10-8 Nb) 4.06 x 10-47 N. The electric force is about 39 powers of ten larger, which makes the gravitational force negligible in the atom.c) Recall a=v2/R and  = v/R.  So, v= 2.2 x 106 m/s and = 4.5 x 1016 rad/s.d) K= 2.3x 10-18 J and L=1.02 x 10-34 kg.m2/s2.e) Since most matter is neutral, electrical forces tend to cancel out as more matter accumulates, and although electrical effects do not entirely disappear, they are effective only at short ranges. Masses cannot neutralize each other so the gravitational force gets very large with very massive objects.

21-4: a) FAB= 3kq2/2d2; FAC= kq2/3d2; FBC= 2kq2/d2. b) FA= 11kq2/6d2 i; FB= 7kq2/2d2  (-i);  FC= 5kq2/3d2 i.d) The only place that this can happen is moving the charge to the right of -q. Setting the forces equal to zero gives you x=4.1d to the right of -q.e) The vertical component of FAB would have to cancel FBC.

21-5:  The answers would be: q1 = 22 C, q2 = ± 127 C, with the charges being of opposite type. The solution would be as follows:The original charges have opposite signs since the force is attractive (negative), this means that Coulomb’s law becomes kq1q2/r2=-25. When they touch the charge is redistributed so each ball has charge q3= (q1+q2)/2 . Since we don't know which charge is negative, q3 could be positive or negative, but either way the force would be repulsive (positive) and Coulomb’s law becomes kq3q3/r2=25. Substitution allows you to relate the two charges with a quadratic equation. Solving the quadratic gives you a relationship between q1 and q2. Plugging back into Coulomb’s law will then give you the value of each charge.

21-6: a) Draw free-body diagrams for one of the charges. The forces (weight, tension and coulomb force) add up to zero and the vector relationship can be used to solve for q.b) (i) the angle of the mass that is increased decreases while the other angle increases, (ii) both angles increase, (iii) the angle of the shorter string increases and the other decreases.c) There is not enough information to solve this problem in general. Assuming that the external electric field is horizontal it is possible to make one ball hang vertically while the other one will rise at angle with the vertical. Then you can find an expression for the external electric field in terms of the other given quantities. If the horizontal external field is E=kqcos(/2)/[2lsin(/2)]2, one sphere will hang vertically and the other will rise to an angle with the vertical.

21-7: a) EA= 11kq/18d2 i; EB= 7kq/4d2  i;  EC= 5kq/3d2 (-i).b) The electric field at the location of the removed sphere would not change but the field at all other points would be changed.

21-9: (a) This was done in class, check your notes. Answer is Ex=2kqa/(x2 +a2)3/2 (-j).b) For y>a, Ey=4kqay/(y2 -a2)2 (j). For y<a, Ey=2kq(y2 +a2)/(a2 -y2)2 (-j). c) In either (a) or (b), the "a" term in the denominator is negligible.d) Along the x-axis the maximum value is at the origin. In the y-direction, the point charge expression predicts an infinite field at +a. Because infinite fields are not realistic the implication is that there really are no such things as "point charges", all charges must have some dimension even if it is too small to measure.

21-10:a) Very similar problem to the one before except vector addition is in the x-direction. Answer is Ex=2kqx/(x2 +a2)3/2 (i).b) For y<a, Ey=4kqay/(a2 -y2)2 (-j). For y>a, Ey=2kq(y2 +a2)/(y2 -a2)2 (j). c) The extreme case is actually proportional to an inverse squared function of distance. This makes sense because this arrangement resembles a point charge from far awayd) To determine max value take a derivative of  Ex and set it to zero (dE/dx =0).  Ex is a max at x=+ 0.71a

21-11: a) It is somewhat less messy to do this along a horizontal  axis passing through the quadrupole. Centering the x-axis on the -2q charge gives, Ex=2kq(-a4 + 3x2a2)/x2(x2 -a2)2 , where "x" is the distance from the -2q charge along that axis (i). Neglecting a, Ex is proportional to 1/x4.b) The basic idea is that close to the center the dipole and quadrupole fields are larger than the point charge, but they decay much quicker with distance.c) As mentioned elsewhere, with more charges of both types the electric effects substantially drop off at long distances and become effective only at short distances. This is why electric forces are very important in chemistry but mostly negligible in astronomy.

21-12: Draw free body diagrams. To determine torques you must consider a specific axis of rotation. a) Dipole is pushed up and has clockwise torque about its center; charge is pushed down and has counter-clockwise torque about the center of the dipole. b) Dipoles are pushed apart, no torques, but rotational equilibrium is unstable. c) Charge and dipole are attracted to each other, no torques. d) One dipole is pushed down and rotates clockwise about its center; the other is pushed up but also rotates clockwise about its center. This doesn’t violate Angular Momentum Conservation because both dipoles, as a whole, will circle counter-clockwise in the plane of the paper due to the net forces on them.

21-13: a) Net force = 2kQqa/(x2 +a2)3/2 (j).b) Net torque about dipole center is clockwise and equals 2kQqax/(x2 +a2)3/2 (-k).c) Force on Q is the same as on the dipole.d) There is also a torque on Q if you consider the axis at the center of the dipole.

21-14: This problem is identical to the projectile problems that we did in kinematics. The acceleration here is not "g" but rather Fe/m=Eq/m. Since the electric field is uniform the acceleration can be considered constant and we can ignore edge effects, so the old kinematics formulas apply.a) t=L/v and the proton has to start a distance D=at2/2= Eq/m(L/v )2/2 above the plate in order to clear it.b) The final velocity of the proton is vfinal =v i + ( EqL/mv) j.c) Deflection angle is given by the tan= EqL/mv2.d) An electron would get a force in the opposite direction and, since it has less mass, it would have a greater acceleration…e) Plate charge would be negative.

21-15: a) t=2vsinm/Eq b) v c) H=(vsinwhere a=Eq/m d) See 14(d)

21-16: a) Similar problem was done in class, check your notes: Ex= kQ/(x2 – L2/4).b) Ey= kL/y(y2 + L2/4)1/2; Ex= 0. c) Ey= kcos -cos )/y; Ex= ksin -sin )/yd) Ey= ky. e) If y<<L,  ~ 0º and ~180º .

21-17: a) F=Ex= kQq/(x2 – L2/4). b) (x2 – L2/4)1/2

21-18: a) Ey= 2k/R.b) Ey= ksin/R; Ey= kcos)/R.; Etotal= 2ksin)/R, directed (180o +), along the bisector of the arc. c)

a) The symmetry of the ring causes the y-components of the electric field to cancel along the axis. A vector diagram would show this very well. bEx= kQx/(x2 +R2)3/2 (i). c) x=0.71Rd) For x<<R, the force on the -q charge would be proportional to "x" and would always be directed toward the center of the ring which makes this a restoring force that leads to SHM.e) The angular frequency constant here would be = 2/T = (kQq/mR3)1/2.

21-20:This was partially done in class. Check you notes. The idea is to use the electric field derived previously along the axis of a ring of charge, then use this expression to integrate rings of variable sizes to get the expression for the entire plate. In this case dQ is the charge of one infinitesimal thin ring of arbitrary radius "r" and thickness "dr" imbedded in the plate. a) E= 2k[1- x/(x2+r2)1/2]b) For x<<R, E=2k, which is independent of distance. This is an important case which tells us that it is possible to experience a constant E field if we are close to a plate of uniform charge.c) For x>>R you can show that the plate acts like a point charge by using the binomial expansion (1+x)p = 1+px, for x<<1.

21-21: a) Torque about center of dipole=Eqdsin c) The dipole is in equilibrium when parallel to the electric field.d) Use the fact that =I and use the small angle approximation as we did for pendulums in mechanics.

21-22: a) F=(kQ2/L2) ln[D2/(D2-L2)]

These are concentric spheres

+q

-2q

+q -q

+Q

-Q

+Q+q+q

See textbook

+q

-2q

See textbook

q1

q2

q1q2 q2

q2q1

Chapter 22:Gauss’ Law Hints Last update 9/10/08

1. A number of these can be seen in the textbook (p 724-5) or they were done in class. Lines-of-force cannot cross in space or bend at sharp angles because this would imply that the electric field had

two different directions at the bend or cross point, which is not possible. Lines-of-force are always perpendicular to charged surfaces because any surface looks flat and infinite if you are

close enough to it, so all lines-of-force begin parallel and perpendicular to the source charged surface before it spreads out in space.

-Charged particles do not follow a path along a single curved line-of-force. The force on the particle is tangent to a line-of-force but as the particle accelerates its velocity is not parallel to the force it experiences at a particular point and the particle follows a curved path that is different from a single line-of-force curve.

2. Since “q” is proportional to the lines emanating from the charge, compare the lines connected to each charged object. Answers: a) q2 =q1=+6nC; b) q2 = -q1=+6nC ; c) q2 = -3q1= -18nC

3. (a) Since E= flux/area, the argument is based on the way that the lines spread out in space. For a point charge the lines-of-force (electric flux) spread into spherical surfaces (E=flux/4r2), so E is prop to 1/r2. For a line of charge the flux spreads into cylindrical

E

E

+3q-2q

+qAB

E

y=a

x=a

z

dA

surfaces (E=flux/2rL), so E is prop to 1/r. For a plate of charge the areas over which the flux spreads do not grow in size (E=flux/A), so the lines-of-force are parallel and E is therefore uniform. The last two arguments are only true if you are away from the ends of the line or plate of charge.

b)The electric field E can be defined as flux/area, since flux is the correct expression of the lines-of-force, one could say that E is the density of the lines-of-force.

c) The main limitation is that the number of lines is finite whereas there is electric field at every point in space.

4.

Illustration Given information Find a) E= 200 N/C; triangular

dimensions are 30 cm,40 cm and 50 cm and the width is 20 cm.

This was done in class: Through slanted surface and back vertical surface, = +12 Nm2/C, all others =0

b) E= 200 N/C; diameter of hemisphere is 40 cm.

Find the flux thru circular base of the hemisphere: = ER2 = +25 Nm2/C

c) Q=2 nC at the center of a cube with sides equal 60 cm.

Because the charge is at the center, it is easier here to use Gauss’ law to determine the entire flux emanating from the charge and then divide by 6: side= 4kQ/6 = +38 Nm2/C

e) q= 2 nC Because these are closed surfaces use Gauss’ law: A= 4k(+q-2q)= -226 Nm2/C; B= 4k(+q+3q)= +905 Nm2/C

f) E is not uniform but it is given by the formula: E=Eoy/a (k^), where a and Eo are constants.

The element of integration here is a strip parallel to the x-axis, a long and dy wide (dA=ady k). Then A= S EdA= So

a (Eoy/a) ady= Eoa2/2

5. (a) Charged spheres require spherical gaussians; rods require cylindrical guassians; and plates require a gaussian surface that have sides that are parallel and perpendicular to the plate.

b) All these charge arrangements have fields that are not uniform enough for a simple gaussian surface.

6. Check your notes as some of these would have been done in class.

Example Line-of- force illustration & gaussian surfaces

Electric field expressions from Gauss’ law

Graph of E vs. r Follow up...

E, dA

dA

E

dA

E

dA

E

a) Point charge “q” E=kq/r2 r^ Is a point charge likely to exist? No...

b) Conducting sphere of charge Q with radius R

for r<R

E=0

for r>R

E=kq/r2 r^

What happens to the field at a point outside if sphere shrinks to 0? Nothing..

c) Non-conducting sphere with volume charge density with inner and outer radii Rin & Rout.

Ri <r<Ro

E=4k(r-Ri

3/r2)/3

r>Ro

E=4k(Ro

3- Ri

3)/3r2

Consider the special case of Rin=0.Ein =4kr/3Eout =4kRo

3/3r2

d) Non-conducting sphere with non-uniform charge distribution =Ar and radius R

r<R

E=kAr2r^

r>R

E=kA R4/r2

What is the net charge?Q=AR4

7. These would have been discussed in class: (a) Since all the charge is on the surface in a conductor, a gaussian surface inside enclosed no charge.

8. Some of these were done in class, so check your notes for hints on how to get the answers. Gaussian surfaces will be spheres.

Example Line-of- force illustration & gaussian surfaces

Electric field expressions Graph of E vs. r

Follow-up...

a) Inner metal sphere (+q) and outer shell of negligible thickness (-q)

for r<a

E=0

a to b

E= kq/r2 r^

for r>b

E=0

inner = -qouter = 0

b) Point (+q) charge at the center of a solid, un-charged conducting shell (radius a to b)

E= kq/r2 r^ E=0 E= kq/r2 r^ inner = -qouter = +q

c) Non-conducting sphere with uniformly distributed (–q) surrounded by a thin conducting shell with (+2q)

E=-kqr/R3 r^

E=-kq/r2 r^ E= kq/r2 r^ inner = +qouter = +q

d) Metal sphere (q1, radius a) surrounded by a thick, metal shell with (q2, b to c)

(a to b)E= kq1/r2

(b to c)E=0

E for r>cE=k(q1+q2)/r2

Varies.. inner = -q1

outer = (q1+q2)

9. Gaussian surfaces here will always be cylinders of arbitrary length and radius where the E field is being determined.

Example Line-of- force illustration & gaussian surfaces

Electric field expressions, measured from the center of the rod.

Graph of E vs r Follow-up...

a) Line of constant charge density dq/dL

E=2k/r r^ Is it possible to have a line of charge? No…

b) Conducting rod of charge Q, length L, and cross-sectional radius R.

for r<R

E=0

for r>R

E=2kQ/Lr

E=2k/r E=4kR/r

c) Line of charge with density surrounded by cylindrical shell with - density and radius R

for r<R

E=2k/r r^

for r>R

E=0

= -/2R on the inside surface and “0” on the outside surface

d) Non-conducting rod with uniform charge distribution =dQ/dV and radius R

r<RE=2kr

r>RE=2kR2/r

You would have to integrate to get qin: qin = SCr(2r ldr)=C2 lr3/3

10. For a numerical example, consider a cylindrical conductor of length 20 cm and diameter 1 cm. A proton 1 cm above the surface of the cylinder orbits the cylinder with a speed of 106 m/s. Determine: (a) the linear charge density of the cylinder, and (b) the net charge of the cylinder. What if the proton’s orbit were 2 cm?

This requires that you review circular motion. The proton’s centripetal force (mv2/r) is provided here by the electric force from the field (Eq). The E field can then be related to the charge density of the rod.

My answers: a) -0.58C/m; b) -0.116 C; c) The radius of the orbit doesn’t change any of the answers because they are independent of “r”.

11. Gaussian surfaces can be cylinders of arbitrary height and crossection.

Example Line-of- force illustration & gaussian surfaces

Electric field expressions, measured from the center of the plate.

Graph of E vs. y

Follow-up...

a

a

E

a

E

ya

a

E

a) Non-conducting sheet of charge with charge Q and area A

E=2kQ =dQ/dA =Q/A

b) Conducting sheet of charge with charge Q, surface area A, and thickness d.

for y< d/2E=0

for y> d/2E=2kQ

’=dq/dA=Q/2A

c) Non-conducting slab of area A, thickness d, with charge Q uniformly distributed.

for y< d/2E=Qydo;Where y is measured from center of slab.

for y> d/2E=Qo

E=yo;

12. In 11a above, Ea = /2o . In 11b above, Ea =’/o . These two expressions appear different because = 2’. But the fields must be equal since the overall charge density is the same.

13. Seen from the outside the slab has an overall area charge density Q/A=, the same as for the sheet of charge of negligible thickness. This is reasonable because outside the slab, the thickness is irrelevant. In all cases in problem 11 the outside field depends only on the overall charge density.

14. The most uniform electric fields are generated by two or more sheets of charge parallel to each other.

Example Line-of- force illustration & gaussian surfaces

Electric field expressions, in between and outside the plates

Graph of E vs. x, form the left-most plate.

Follow-up...

b) Two conducting sheets of area A with equal and opposite charges +Q & -Q

outsideE=0

betweenE=Qo

E= o

c) Two conducting sheets of area A with charges +Q & -2Q

E=Qo;Toward the plates

E=3Qo;from + to -

-Q/2 and +3Q/2 on +plate and -3Q/2 and –Q/2 on – plate.

d) Three sheets of area A with charges +Q, +Q, and -Q

E=Qo;Away from the plates

E=Qo;From 1st to 2nd plate.E=3Qo

from 2nd to 3rd

plate.

+Q/2 & +Q/2 on 1st plate; -Q/2 and +3Q/2 on 2nd plate; -3Q/2& +Q/2 on 3rd plate.

a

q,m

l

q

-Note: From afar the net effect of many plates is that of one plate with a net charge, the field must emanate equally in both direction because of the symmetry of space so the outside fields must be equal and opposite.

15. This is numerical version of 14b. If you follow similar procedures you can get the formulas for the fields outside and in between the plates.

Use 2 Gaussian surfaces to relate the field between the plates and the field outside the plates. Use the fact that by symmetry the fields outside must be equal and opposite.

Answers: a) Eo=1.2 x106 N/C & Ei=4.2 x106 N/C; b) o=10.6 C/m2 & i=36.9 C/m2; c) qo=0.85 C & Ei=2.95 C; d) Determine the acceleration of the electron and use the appropriate kinematics formula.

16. As a review of simple harmonic motion consider this problem. A small object of mass

m and charge q is attached by a thread of length l to a large, flat, non-conducting sheet of

charge with area charge density and same sign as q. If the object is displaced slightly

from equilibrium, show that it will undergo SHM with a period of T= 2(2oml/q)1/2.

Solution: This is basically a simple pendulum. Recall that the period of a simple pendulum is T=2(l/g)1/2. Here the “g” is by analogy the acceleration due to the electric field, which is a=F/m=Eq/m. Finally the electric field here is due to a flat sheet of charge so that E=o. After substituting you get the expression given.

17. Gauss’ law can also be applied to gravitation by defining analogous quantities to those in electricity.

a) Show that the gravitational acceleration “g” can be defined as the “gravitational field” at a point.b) Define “gravitational flux” and use it to express Gauss’ law for gravity.c) Use Gauss’s law to prove that the gravitational field outside a planet with uniform mass density is the same as if the mass

were concentrated at the center. (Newton had to use calculus to prove this, since he didn’t have Gauss’ law).a) By analogy to the electric field (F/q) the gravitational field “Eg” should be F/m. Since F=GMm/r2, Eg= GM/r2 which is “g, the acceleration due to gravity.

b) Gravitational flux should be g= Eg dA and Gauss’ law of gravity should be Eg dA= 4GMinside.

c) You can use Gauss’ law here the same way we used it to show that sphere of uniform charge had the same field as a point charge outside.

18. After J.J. Thomson discovered that atoms contained negative particles (now called electrons). He postulated the first atomic model that consisted of electrons distributed throughout a uniform ball of positive charged material (protons and the nucleus had not yet been discovered). He described this model as plums in a pudding hence it came to be known as the “plum-pudding” model of the atom. In this model the electrons did not “orbit” but were located at “equilibrium positions” inside the atom.

a) Explain why, in the hydrogen atom, the equilibrium position of the electron is at the center of the mass of positive matter. That’s where the field is zero…

b) Show that if the electron is displaced by a small amount from the center, the resulting motion is simple harmonic. The field from the positive distribution depends on “r” so there would be a restoring force proportional to the position of the electron, suggesting a SHM case…

c) In Thomson’s time it was known that atoms emit light waves when excited. Thomson postulated that this was caused by the electron vibrating back and forth. What would be the size of the hydrogen atom in this model to produce light of frequency 4.5 x 1014 Hz? Compare to the known size of the hydrogen atom of about 10-10 m. Hint: Use the period derived from the SHM in (b).

d) If the electron were displaced from equilibrium by a distance greater than the radius of the positive ball of charge, would the resulting motion still be simple harmonic? Explain. No…the force becomes inverse squared with distance…

e) Consider a Thomson model of the Helium atom, with two electrons and a positive spherical mass of charge +2e of radius R. How far from the center must the two electrons have to be for the atom for be stable? Hint: Locate the point where the electric field from the +charge mass and the other –electron would cancel.

19. Thomson’s model of the atom was proven incorrect when Rutherford discovered that the positive “stuff” (what we now call the protons) was concentrated inside a tiny region of the atom (about 10-14 m in diameter). The modern “quantum model” describes the hydrogen atom as consisting of a “point” positive +e charge at the center and the electron “smeared out” over the

volume surrounding the nucleus. The electron charge density is given by the equation:

ρ(r )=−( Qπao

3 )e−2 r/ao

, where ao= 5.3 x 10-11 m and is called the Bohr radius, and Q is the elementary charge “e”= 1.6 x 10-19C. This is a very challenging problem.

a) Find the total amount of charge within a volume of radius “r”. Show that as rinfinity, the net enclosed charge is zero. Explain that result.

The first thing is to find the amount of charge due to the “negative cloud” created by the electron. This requires that you integrate the given density function over a sphere of radius “r” within the volume.

q(r)= dV= r2dr)= (-Q/ao3) e-2r/a r2dr)=….

The solution can be found in a “Table of Integrals”, in general x2e-bx dx= -(1/b3)(b2x2+ 2bx+2)e-bx

b) Find the electric field caused by the hydrogen atom as a function of r.c) Graph the electric field vs. r.

Chapter 23: Electric Potential Hints & Answer- Updated 9/19/08

1.

a) Fill in the following chart: Ug stands for gravitational potential energy and Wg is the work done by the force of gravity. Remember that potential energy is relative to an arbitrarily chosen U=0.

Reference Ug A Ug B Ug C UAB UAC UBC Wg AB Wg AC Wg BC

Relative to table-top

0 0 30 J 0 30 J 30 J 0 -30 J -30 J

Relative to the floor

20 J 20 J 50J 0 30 J 30 J 0 -30 J -30 J

Relative to the ceiling

-40J -40J -10J 0 30 J 30 J 0 -30 J -30 J

b) This chart illustrates the definition of potential energy because it shows that U= - Wg, which is the definition of potential energy and potential energy requires a conservative force.

c) The work done by and against gravity depends only of vertical displacement.

E

C

Aa

B 0.04 m

d) K= - U= 50 J = mv2/2.

2. The following is an electric version of the problem above. A +2 C charge is moved from point A to point B then to point C inside a uniform electric field E=200 N/C. The distance from Cto B is 4 cm. The angle is 370.

CA B

a) & b)Fill in the following chart. Here Ue represents the electric potential energy and We the work done by the electric force. The symbol UAB represents the change in U from A to B = UB UA.

Reference Ue A Ue B Ue C UAB UAC UBC We AB We AC We BC

Relative to A 0; 0 0; 0 +16J;-24J

0; 0 +16J;-24J

+16J;-24J

0; 0 -16J;+24J

-16J;+24J

Relative to B “ “ “ “ “ “ “ “ “Relative to C -16J;

+24J-16J;+24J

0; 0 “ “ “ “ “ “

c) The potential differences are: VAB=0, VAC=8 v, and VBC=8 v.

d) You are not encumbered by the variations in + and –energies that result from the two different types of charge that can move in an electric field. Like electric field, electric potential describes the energy properties of space itself.

3. Enough positive and equal negative charge is accumulated on two flat parallel plates until the voltage between them is 36 volts. Define the negative plate as zero volts (that is, “ground”).

a) The broken lines drawn parallel to the plates are called equipotentials because being perpendicular to the electric field the potential cannot change along these lines. Plates are themselves equipotentials for the same reason. Here E is constant so V is proportional to s and the lines must be equally spaced.

b) VA= 6v, VB= 12v, etc....(V= 6v)c) VAB= 6v, VAE= 24v, VEC= -12v, & VDB= -12v. Calculate the following:d)

(i) 18J(ii) -6J.(iii) 12e J(iv) -36J & 36J

e) E= 600N/C & = o600 C/m2. f)

g) The answers to b) would differ, they would be negative starting from the positive plate. Also the shape of the graph in (f) would be the same but V=0 at the positive plate and the other potential values would be (-).

4. Consider a configuration of three charged flat plates set parallel to each other. As you go from one plate to the other the potential changes according to the graph shown below the plates.

a) Taking the negative of the slopes of V, the electric fields are zero outside, EAB= -200i & EBC= 1200i.

b) Draw a graph of the electric field vs distance. Remember V decreases in the Direction of the field.

+100v

-200v

0.50m 0.75m

V+1200N/C

-200N/C0.50m 0.75m

E

0.06m

V+36N/C

-12N/C

0.06m

E+

-

EA EB

1 cm

c) To find acceleration, a= eE/m. To find speed use K=Vq = mv2/2. d) The electron will not make it to the other plate because the electric field is too strong. It will be decelerated to v=0 and turned around after moving 1/6 of the way to C.

5. Consider two parallel conducting plates of area 0.028 m2 and different amounts of charge. One plate has a net charge of +6pC

(10-12C) and the other -12pC and they are 6 cm apart.

a) Using Gauss’ law you can show that Eout= ( + )/2 o= 12 N/C . and that Ein= 3Eout=36 N/C.b) Remember V decreases in the direction of the field.

6. Identify the following statements as either true of false. Justify your choice.

a) Equipotential lines are always perpendicular to lines of force. Trueb) Equipotential lines can intersect in space. Falsec) Equipotential lines spread out as the electric field gets weaker. Trued) The electric potential inside a conductor is always zero. Falsee) If you touch a high-voltage wire you will invariably be electrocuted. Falsef) Electrons will freely move toward higher potentials. Trueg) If an electron and a proton are accelerated through the same potential difference, they will have the same acceleration

(False )and they will acquire the same amount of energy (True).

7. The figure shows a number of equipotentials curves in a cross-section of space.

a) At B the equipotentials are denser.b) On diagram.

+Q

-Q+q

-2q

+Q+Q +q+q

+q -q

-Q

-Q

r

-Fg

r

-U

c) The electric field intensity is given approximately by (V/x):at point A is about 500 N/C and at point B is about 1000N/C.

d) The electric field vectors are parallel to the line of force but their directionsdepend on whether the source at the center is positive of negative, and that is not

given in this problem.

8. Since equipotentials are perpendicular to the lines of force they tend to follow the surface contour of the source charge. They are closer together near points and sharp corners because charges accumulate more densely at those locations. Like lines of force they can blend but not cross. Note how the equipotentials follow the “contour” of the charged objects.

9. The potential in a certain region in space is given by V=axy, for x & y >0, and a is a constant.

This is conceptually challenging problem.

a) Ex = dV/dx = -ay; and Ey =dV/dy = -axb) Equipotentials would be a series of hyperbolas with different constant V’s.c) Two equally (-) charged plates at right angles to each other could do something like this.

10. The following problem will review the gravitational potential energy for spherically symmetric gravitational fields such as those from planets and stars. A rocket of mass 105 kg moves from the surface of the earth to a distance of 3 earth-radii above the surface of the earth. Recall that we define the zero of gravitational potential energy at infinity in these cases.

a) Uinit=-GMEm/RE= -6.23 x 1012 J and Ufinal= -GMEm/4RE= -1.56 x 1012 J.b) Ug= +4.67 x 1012 J and Wg = -4.67 x 1012 J.c) Ug= -GMEm/RE (1/8 -1/4) = +0.78 x 1012 J. The reason it is so much less is that further from the earth the force of gravity is

much less therefore energy changes require more distance.d) Graphs:

e) We chose Ug=0 at infinity because it gives us the simplest potential energy function Ug= -GMm/r.

These are concentric spheres

r

-U

r

+V

r

E

R

+V

R

E

r

+V

a b c

+V

a b c

E

R

+V

R

E

11. Now for an electric version of the problem above. When the electron in the hydrogen atom is “excited” it rises from its ground energy state position (r=0.52 x 10-10 m) to a position 4 times further.

a) Uinit= -ke2/R= -4.4 x 10-18 J= -27.7 ev, and Ufinal= - ke2/4R = -1.1 x 10-18 J= -6.9 ev. For potentials divide by the electron charge: Vinit= +27.7 v, and Vfinal= +6.9 v, the + sign here reflects the fact that the potential source is a positive charge.

b) UE= +3.3 x 10-18 J= 20.8 ev and WE = -3.3 x 10-18 J.c) Graphs:

d) The energy values would change sign but not the potential values.e) Same answer as 2d.

12. Consider the following examples of spherically symmetric electric fields. Fill in the following chart summarizing important properties. Some of these were done in the previous problem set. Remember Voo=0.

Example Equipotentials illustration

V V at relevant surfaces

Graph of E vs. r

Graph of V vs. r

a) Point charge +q kq/r2 kq/r

b) Conducting sphere of charge Q with radius R

Eout=kQ/r2

Ein=0

Vout =kQ/r

Vin =kQ/R

Vsurface =kQ/R

c) Charged metal sphere of radius a surrounded by un-charged conducting shell with radii b to c.

Eout=kq/r2

Ebetween= kq/r2

Ein=0

Vout =kq/r

Vbetween =kq[(1/r) +(1/c)-(1/b)]

Vc =kq/c

Vb =kq/c

Va =(kq/c)+Vba

d) Non-conducting sphere with radius R & charge Q uniformly distributed

Eout=kQ/r2

Ein= kQr/R3

Vout =kQ/r

Vin =(3kQ/2R)- (kQr2/2R3)

VR =kQ/R

Vcener =3kQ/2R

13. Two conducting sphere of different radii a & b have equal amounts of charge Q on them.

(d-x)

-q

d

4q

0(x-d)

V=0

r

y

+q

a

-q

-ax

a) Va >Vb.b) Charges will move until the potentials are equal.c) The final charge ratio qa/qb=a/bd) The final surface density ratio a/b =b/a. This shows final surface densities are NOT equal which is a common

misconception.e) If b=2a, then qa/qb=1/2; a/b =2/1. It is also possible to figure out how the original 2Q charge has been divided. Since

qa+qb= 2Q, you can figure out that qa=2Q/3 and that qb=4Q/3.

14.

a) Because potential is not a vector you don’t have to worry about directions. Add up potentials from both charges and set equal to zero to find answers: (4q/x) + q/(d-x) =0, the two possible places are x=4d/3 and x=4d/5.

These are only two points in an equipotential surface that

encircles the –q charge.

b) There is only one point where E=0, and that’s at x=2d. The reason the zero points are not the same is simply that E=-dV/dx and just because a function

has a zero value it doesn’t mean its derivative is also zero. Remember that E

changes in value as you pass over the E=0 point.

15. Energy conservation requires that Uo + Ko= U+K. In this problem the Uo=-kq1q2/d, Ko=0, U=0, and K=m1v12/2 + m2v2

2/2. Momentum conservation requires that po1 + po2= p1 + p2. The particles start at rest, so 0 +0= m1v1 - m2v2.

Using the givens you can combine these equations and get expressions for the final speeds of the particles. I get 7.24 x105 m/s and 4.24 x107 m/s.

16.

a) Vy=0. Vx= kp/(x2-a2) for |x|>|a|. For |x|<|a|,Vx= 2kqx/(a2-x2).b) The electron here has no energy with respect to infinity but it experiences a +i horizontal electric force due to the electric

field. The electron will accelerate and curve toward the x-axis, it will cross the axis with max speed and begin to decelerate and curve toward the (-)y-axis. It should come to a (temporary) rest at the same distance below zero as it started above zero. Then it will retrace the path back to its original location.

c) & d)Begin with the approximation Vxy= kq[(r-acos)-1-(r+acos)-1]. Since a<<r, this expression can be simplified to

Vxy= kq2acos/(r2-a2cos2) = kpx/r3.

e) This is only worth doing if you want to practice your math skills: E x= -dV/dx in addition Ey= -dV/dy, where r=(x2+y2)-1/2. Cranking through the derivatives gives

E= kp

r3[ ( 3cos2θ−1 ) i+3sin θ cosθj ]

y

+q

a

+q

-ax

b

a

a b

+V

r

E

L/2 x-L/2

l

R 1 cm

x

This show that it is possible to get expressions for complex electric fields if you first determine the potential which is easier to do since it is a scalar.

17. Consider a set of two +q charges a distance 2a apart on the x-axis.

a) Vx= 2kqx/(x2-a2) for |x|>|a|.b) Vy= 2kq/(y2+a2)-1/2.c) In both cases a becomes negligible and you get Vx= 2kq/x, Vy= 2kq/y, which are the point potentials for a 2q charge.

18. Another important symmetry to study is the cylindrical spreading electric field. Imagine a long rod with linear charge density and cross-section radius a.

a) You will have to integrate the electric field from a to b: Vab= -S 2kdr/r=2kln(a/b).

b) It doesn’t make sense to define the zero of potential at infinity here because it doesn’t simplify the potential function. If we arbitrarily define the zero potential at b, then V(r) = 2kln(b/r). This means that for r>b, V<0 and for r<b, V>0.

c) Since E= -dV/dr = 2k/r, which is the correct expression for the electric field of a rod.d) Graphs: of V vs r and of E vs r, again taking V=0 at b.

e) The potential between the rod and cylinder would be the same as for the rod alone. Outside, the potential is the same as on the surface of the cylinder since Eout=0.

19. Determine the potential function for a thin rod of charge Q and length L along the x-axis a distance “x” from the center of the rod.

a) Here it is better to integrate the potential contributions of each element of charge dQ: V= S kdl /(x-l) = kln[(x+.5L)/(x-.5L)]

b) Set-up is similar to problem above but resulting integral in not a common one:

V= kCd/(x-) , and you would need a table of integrals to determine expression.

c) Show that the results in (a) above approaches that for a point charge as x>>a, but not in (b) above.

20. Consider a ring of radius R and charge Q.

x

+V

.71R

E

b

a

a b

+V

a b

E

b

a

E

a b

+V

a b

25 cm

10 cmE

a b

+V

a b

a) The integral here is trivial because potential is a scalar, Vx=kQ/(R2 + x2)1/2.b) Again, Ex= -dV/dx= kQx/(R2 + x2)3/2.

c) Obviously V is a maximum at x=0. E is a maximum at x= + 0.71R as derived in the previous chapter.

d) Draw graphs of V vs r and of E vs r.

21. Consider a circular plate of charge Q and radius R.

a) & b) This problem is similar to the one we did to find the field. Divide the plate into infinitesimal rings of charge dQ, radius r, and potential dV=kdQ/(r2 + x2)1/2. Using the charge density dQ=2rdr. It is fairly easy to integrate dV and get V= 2kQ[(R2 + x2)1/2-x]/R2.

22. Some of the more “tricky” problems involve concentric spheres carrying different charges. Consider the following cases. Assume both sphere and shell are conductors and that the shell’s thickness is negligible.

Example Equipotential illustration

Vba=Vb–Va V(r) V at a and at b w/infinity

Graph of E vs. r

Graph of V vs. r

a) Inner sphere charge is +q and outer sphere of -q

Equipotentials are all spherical surfaces..

Vba = kq(1/a -1/b)

Vout =0;

Vin= kq(1/r -1/b)

Vb=0;

Va= kq(1/a -1/b)

b) Inner sphere charge is +q and outer sphere of +q

Vba = kq(1/a -1/b)

Vout =k2q/r;

Vin= kq(1/r +1/b)

Vb =k2q/b;

Va= kq(1/a + 1/b)

c) Inner sphere charge is -2nC and outer sphere of +4nC

Vba = kq(1/a -1/b) = -108 v

Vout =18/r;

Vin= 144 - (18/r)

Vb =72 v;

Va= -36 v

b

a

a

b

y

x

Vad, Vd

vxLD

d) Inner sphere charge is q1 and outer sphere of q2

Vba = kq1(1/a -1/b)

Vout = k(q1+ q2)/r;

Vin= k(q1/r + q2/b)

Vb = k(q1+ q2)/b;

Va= k(q1/a+ q2/b)

Depends on charges

Depends on charges

23. This change does not affect the potentials outside the shell, but the potential difference between inner and outer spheres will be less than before and the potential of the inner sphere will be less also: Vb =72 v; Vba =

-90 v; Va = -18 v.

23.5 Assuming a vertical orientation of the plate, E= (x/o)i, where x is the distance from the center of the plate perpendicular to the surface. Integrating the field inside the plate gives V=-(D2/2o). The potential decreases parabolically from the center.

24. a) For any two point a & b in an electric field, ∮E⋅ds=V ab+V ba=V ab−V ab=0

, qed.

b) Start out by tracing out a rectangular closed path in the electric field that goes parallel

and perpendicular to the lines of force as shown. You can argue that the integral on this

closed path cannot be zero because the field is stronger on the upper part of the path. This

contradicts the requirement that the potential be path independent (conservative) so a field

that increases in magnitude without changing direction is not possible.

25. This is mostly a kinematics problem. The electron spends time t1= L/vx between the plates accelerating toward the positive plate at rate a=eE/m= eVd/md. It travels vertically y1=at1

2/2, and comes out with a vertical component of velocity vy=at1. After exiting the plates, the electron has no acceleration but it spends time t2= D/vx moving toward the screen and traveling vertically and additional y2= vy t2.

a) The final vertical position of the beam when it strikes the screen is Y=y1 +y2. Putting everything in terms of the given quantities you get, Y= (eVdL/mdvx

2)(L/2 + D), which is proportional to Vd.b) Because the vertical position of the beam is proportional to Vd of the plates, one can connect an external potential to the

plates and measure it with the beam displacement.c) The horizontal sweep is accomplished by having a second set of parallel plates perpendicular to the first that deflects the

beam back and forth at a particular rate.

Chapter 24: Capacitance Hints & Answers-Updated 10/4/08

V(volts)2.0

1.5

1.0

0.5

0.0

4 8 12 16 q(mC)

1. A useful approach to learning a new concept is to find and analogy. Consider the following analogies to a capacitor: i) an air tank; ii) a spring. Justify your answers to the questions below.

a) Analogies to the stored charge are: air mass for the tank; stretch or compression for the spring.b) Analogies to the voltage are: pressure for the tank; spring force for the spring.c) Analogies to the capacitance are: mass per pressure; (stretch/force) = 1/(spring constant “k”).

2. You should realize that the “work to create”=potential energy stored:

Example Illustration Work done by an external agent to create arrangement

Potential Energy stored

a) Four point charges, +q, -q, -2q, & +3q, are at the corners of a square with side b

Positions affect the answer. For clockwise order of the listed charges: U = kqiqj/rij = -5.5kq2/b

U=Work against FE = -Work by FE

“U” is always equal to the work done by an external agent to create something against a conservative force, such as the electric force.

b) A water molecule H2O where the distance between the O and H atoms is 10-10 m and the angle is 1050

Recall that H ions are +e and O ion are -2e, then: U = -7.8x10-18 joules = -49 ev

U =Work against FE

The electric force would be doing work of opposite sign to the work done by external agent and the change in U.

c) Conducting sphere of charge Q with radius R

Use:U=k ∫qdq/r= kQ2/2R

U=Work.

d) Two parallel plates of area A separated a distance d with charges +q and -q

Use:U=∫qdq/C=q2d/2Ao

U=Work.

3.

a) K= -Uq= (Vat+q)q= (0.59kq/b)(q)=0.59kq2/bb) Work= U= [k(2Q)2/2R]- [kQ2/2R]= 3kQ2/2Rc) Using U= q2/2C; Work= U= (2dq2/2oA) - (dq2/2oA)= dq2/2oAd) This lowers the stored energy of the molecule since the electrons spend some of the time closer to the hydrogen atoms

than is suggested by the arrangement given in (b) above.

4.

a) Do your best to estimate a max slope from the graph and a min slope, then C=1/slope. I get roughly Cmin=7.5 mF and Cmax=8.5 mF.

b) Implied uncertainty = (Cmax - Cmin)/2 = + 0.5 mFc) Would have a lower slope.d) The energy stored in the capacitor.e) It would mean that the capacitance was not a constant.

5. Some of these were done in class.

VsourceC2 C1

VsourceC2

C1

a) Co= oA/d, with dielectric in between plates C= oA/d,b) C= R/kc) Recall that Vba=kq(b-a)/ab for concentric sphere, then C=q/V=ab/k(b-a)d) For concentric cylinders Vba=2kln(b/a), so C=L/2kln(b/a).e) In this case (b-a) d and aba2. The concentric spheres behave like two parallel plates of area=4 a2 and separation (b-a).

9.

a) Since the separation between cans is small I used a parallel plate approximation: C= oA/d. Using =5 for glass and the given dimensions I got C 311pF.

b) The two connected plates are at the same (-) potential while the third plate between them accumulates + charge at the top and the bottom. This arrangement is effectively two parallel plate capacitors in parallel so C= 2oA/d.

c) Since each foil layer in the rolled-up capacitor "sees" an oppositely charged layer on both sides, they store charge on both sides of the foil, so that, as in (b), C= 2oA/d . Using the given dimensions and 2.3 for polyethylene, I get C=0.26F.

7. Using U= CV2/2, the Leyden jar stores more energy: Uleyden= 2.5 J while Ucap= 0.033 J

8.

9. (a) Conceptually one can see that since the capacitors share the voltage, neither can store as much charge as if they were directly connected to a battery. In addition since they have the same amount of charge the smaller capacitor needs more voltage to store that charge; and/or (b) proof is in 8 above.

Parallel Set-up Series Set-upq V U q V U

C1 C1Vs Vs C1Vs2/2 C1C2Vs/

(C1+C2)C2Vs/(C1+C2)

(C1/2)*[C2Vs/(C1+C2)]2

C2 C1Vs Vs C2Vs2/2 C1C2Vs/

(C1+C2)C1Vs/(C1+C2)

(C2/2)*[C1Vs/(C1+C2)]2

Ceq (C1+C2)Vs Vs (C1+C2)*Vs

2/2

C1C2Vs/(C1+C2)

Vs C1C2Vs2/

2(C1+C2)

C C C

ba

C C C

baC C C

ba

10. First Ceq= C/3. In the second, note that two of the capacitors are shorted, Ceq= C, not. In the third each capacitor has the same voltage ab, Ceq= 3C

11. I) Ceq=3F. Individual caps: 4F: +72C, 18v; 8F: +48C, 6v; 4F: +24C, 6v. Equipotential at c =6v.

II) Ceq=1.2 F. Individual caps: 3F: +29C, 9.6v; 2F: +19C, 9.6v; 1F: +9.6C, 9.6v; 6F: +29C, 4.8v. Equipotential at d =4.8v and equipotential at c =14.4v.

III) Ceq=2 F. Individual caps: 3F: +24C, 8v; 6F: +24C, 4v; 2F: +24C, 12v; 4F: +48C, 12v. Equipotential at d =20v and equipotential at c =12v.

12.

a) Max Vab= 40 v.b) The energy in the 6 F capacitor decrease because its potential will decrease as the potential across the 2 F increases. The

energy of the entire circuit decrease as the Ceq decreases.c) This turns out to be a “trick question”, there is no possible way to do this by replacing the 6F capacitor. Since this

capacitor is in series with a 3F, the combination in that branch can never exceed 3F no matter how larger is the other capacitor. The greatest possible Ceq here is 2.1F, but you would need 10 times more capacitance to store 6 mJ of energy with 24 v.

13. (a) The common voltage is 64 v. (b) You would get that the energy decreases in the process which suggests that some energy is being lost to heating in the wires while the charges flow during rearrangement. (c) The common voltage is much less here: 7.1 v.

14. The potentials are Va= 12 v and Va= 24 v with the switch open When the switch is closed potential equalize to Vab= 18 v. In the process +54 C flowed from b to a through the switch as the charges rearranged themselves going from two pairs of capacitors in series to two pairs in parallel.

12 vC2 C1

12 vC2

C1

H

h K

K

x

L

d

y K

15. Here is a good time to go to the text and study some of the illustration and maybe even read a little of the text. The “free” charge comes from the source and is free to return to the source, whereas the “bound” charge is the induced charge on the dielectric. The voltage is related to the net electric field between the plates which are the result of the sum of the “free” and “bound” charge.

16. a) I’ve added voltage values under the charge values in each capacitor.

b)

17. I: Ck= Co / [1- y(-1)/d] II: Ck= Co [1+ x(-1)/L] III: Ck= Co [1+ h(-1)/H]

IV: Ck= 3Co

C1=3F C2=6F Cequivalent

q1(C) U1(J) q2 U2 qtotal Utotal

Battery connected, no dielectric (V=12 v)

+36(12v)

216 +72(12v)

432 +108(12v)

648

Battery connected, K=2 dielectric introduced into C1

+72(12v)

432 +72(12v)

432 +144(12v)

864

Battery disconnected, V=16v dielectric K=2 removed

+48(16v)

384 +96(16v)

768 +144(16v)

1152

C1=3F C2=6F Cequivalent

q1(C) U1(J) q2 U2 qtotal Utotal

Battery connected, no dielectric Ceq =2F

+24(8v)

96 +24(4v)

48 +24(12v)

144

Battery connected, K=2 dielectric into C1(Ceq =3F)

+36(6v)

108 +36(6v)

108 +36(12v)

216

Battery disconnected, dielectric K=2 removed

+36(12v)

216 +36(6v)

108 +36(18v)

324

a) With a conductor E=0 (infinity) and the equivalent is one narrower capacitor with plates separated a distance (d-y). Ceq = Co d/(d-y)

b) If you place the dielectric midway inside the capacitor you would have three separate capacitor segments but the results are the same as in (a).

c) Keq=3

18. Since the dielectric is attracted to the plates, an external agent would do –Work in introducing the dielectric into the capacitor. In this problem the battery is disconnected so the charge Qo = (Co Vo) is a constant, but the voltage decreases.

(a) From problem 17 above, Ck= Co [1+ x(-1)/L]

(b) Note that the potential energy decreases as the dielectric is introduced. In terms of the original voltage and capacitances: Uk= Qo

2 /2Ck = Co2

Vo2 /2Ck . In addition the voltage decreases as dielectric is introduced so the instantaneous voltage is Vk = Qo /Ck = (Co

Vo)/Ck .

(c) F=-dU/dx=[Co3

Vo2 /2Ck 2](k-1)/L= Co Vo

2 (-1)/2L[1+ x(-1)/L]2= Co Vk2(-1)/2L, where Vk is the instantaneous voltage.

19. (a) Ck is independent of voltage or charge so it is the same as in the previous problem.

(b) In this case the voltage remains constant and the charge increases: Uk= Ck Vo2/2.

(c) This problem cannot be solved by taking the derivative (-dU/dx) as in the previous problem because the battery does +work adding potential energy to the system. Rather you must use the results from the previous problem since ultimately the force depends on the particular charge and voltage at the position x of the dielectric. For an answer rewrite the solution to F above in terms of the instantaneous voltage Vk which turns out to be F= Co Vk

2(-1)/2L. Here Vk= Vo, so F= Co Vo2(-1)/2L

20. a) Integrate from R to infinity; U=S (o E2/2)d(volume)= (ok2q2 /2) S4r2dr/r4 = kQ2/2R.

b) Here you want to integrate the field for a non-conducting sphere (E=kQr/R3) from 0 to R, so U= kQ2/10R, which is 5Uout.

21. a) Ux= xq2/2o A, where x is the instantaneous separation between plates. You can then conclude that at twice the separation you have twice the energy.

b) F=-dU/dx=-q2/2o A. This equals F=-qE/2, where E=q/o A for in-between plates.

c) You might expect F=q from the definition of the electric field. The reason the force is half what you might expect, is that each plate only experience the field from the other plate and not the field it generates, so the field acting on each plate is half the field in between plates.

Chapter 25: Current & Resistance Hints & Answers-Updated 10/9/08

1. Answer the following questions. Always explain your reasoning.

a) No we have a source of voltage so that a potential difference can be maintained across a conductor.b) The voltage source sets up an electric field inside the conductor that sets all the free charges in motion in a very short

time…c) The charges accelerate between collisions according to Newton’s 2nd law, but they are constantly colliding so on average

they move with a constant drift velocity.d) No, there are non-ohmic regions in conductors, diodes, etc…e) The textbook has a pretty good discussion on this..f) In N-type semiconductors the predominant charge carriers are electrons (-) while in P-type semiconductors the

predominant charge carriers are holes (+)…

2. a) 30 persons.

b) people flow rate= people/time= (car speed)(cars/distance)(# of people in cars)(# of lanes)

c) people/timeI; n(cars/distance); q(# of people in cars); A(# of lanes); vd(car speed).

3. Basically you need to plug in the given data. In the salt solution there are double the number of charge carries because the flow of +ions effectively doubles the current.

a) Current density J=I/A= I(4/D2)b) Wire: vd=0.72 mm/s; solution: vd= 6.5 mm/s; beam: vd=3.6 x 107 m/s.

4. a) 8.9(g/cm3)x(1mole Cu/63.55 g)x((6.02x1023 atoms/mole)x(106cm3/m3) = 0.84 x1029 atoms/m3, which compares fairly well with the given value of 1.1 x1029 atoms/m3. The comparison suggests there is a bit more than one (1.3) charge carrier per atom of copper.

b) Plugging into the formula I get 253 oC.

c) You can rewrite the expression given in (b) as T. For infinitesimal small changes ddT. Rewrite as an integral: ∫ (d.∫dT and solve…

5.

Change in resistor Illustration New resistance Ratio: Rnew/Roriginal

a) The length is halved and the diameter is doubled.

R’=(L/2)/4A 1/8

b) The wire is hollowed out at the center to half its radius.

R’=L/(3A/4) 4/3

c) The wire is cut in two and the pieces are glued along their lengths.

R’=(L/2)/2A 1/4

d) The wire is stretched until is length is doubled without changing its mass.

R’=2L/(A/2) 4

A B

V (volt)

I (amp)

3.88 0

I (amp)

0V (volt)

R(ohm)

0I (amp)

V (volt)

I (amp)

3.88 0

I (amp)

0

V (volt)

I (amp)

0

V (volt)

6. Consider two cylindrical conductors connected end to end and connected to a battery of voltage V at the free ends. The conductors have the same cross-section but conductor A is twice as long and has three times the resistivity of conductor B.

a) IA/IB=1/1; RA/RB=6/1; VA/VB=6/1; EA/EB=3/1; PA/PB=6/1.b) VA = 6V/7; VB=V/7; and the potential at the junction is V/7 relative to ground.c) Note the Gaussian surface in the illustration. The arrows indicate the electric field and “a” is the cross-sectional area.

Using Gauss’ law: -aEA + aEB=ao = -a3EB + aEB o2EB = o2V/7L . 7.

8. Compare the following three graphs that describe the I vs. V relationship for three different devices: a conducting wire, a vacuum diode, and a semiconductor diode.

Nichrome resistor Thyrite resistorData point 1 2 3 4 1 2 3 3

I (A) 0.5 1.00 2.00 4.00 0.5 1.00 2.00 4.00

V (v) 1.94 3.88 7.76 15.52 2.55 3.11 3.77 4.58

Average R 3.88 3.88 3.88 3.88 5.10 1.12 0.66 0.41

,

y K

J=0.10 A/cm2y

x

a) In one direction they all have some region where I increases with V and, near the origin, the relationship could be considered “ohmic”. The first two flatten out with increased voltage, though.

b) The diodes behave very differently from the conductor in the reverse voltage. The two diodes behave differently at high voltages, the vacuum diode “saturates”, whereas the semiconductor diode becomes more conductive and eventually “breaks down”.

c) The don’t allow current flow with a reverse voltage…d) Review discussion about resistivities in the text, though you may not get as much information as on the web. Briefly,

saturation in conductors occur when increasing temperatures lead to more and more electron collisions and the drift velocity of the electrons cannot increase. In vacuum diodes saturation occurs when the number of electrons jumping the gap reach the maximum possible. The semiconductor diodes break down because currents cannot increase indefinitely without fundamentally changing the properties of the material.

9. Check out the textbook, or other textbooks, or the web…The most common use for diodes is to keep current flowing in one direction.

10. Start with the definition of power: U/t=qV/t=Power=VI=I2R=V2/R. Check your notes for details.

11.

a) Neglecting the voltage drop across the wires, I=P/V=0.522 A b) Rcord=0.0642 ohm, and Pcord=0.0175 watt, obviously ignoring the cord is a good approximation.c) The 40-watt light bulb (R=330 ohms) has more resistance than the 60-watt one (R=220 ohms).d) The “40-watt” light bulb would be brighter but it would be generating less power than before. If you assume the

resistances are the same (which is not correct since these are non-ohmic devices) it would only be a 14.4-watt bulb as opposed to the “60-watt” bulb which would only be 9.6-watts.

12. Note that the given 115 kV-voltage is not the voltage across the transmission cable but between the cable and the ground (this is called the “line voltage”). The resistance for the entire 10 km of cable is 0.50 ohm.

a) I=P/V=8.7 kA; Vcable = IRcable= 4350 volts, which is small compared to the “line voltage” of 115 kv.

b) Ploss=I2Rcable = 37.8 MW which is 3.78 % of the 1000 MW plant output.

c) You want the Ploss in (b) to equal 20 MW and that would require reducing the current through the cable to 6.3 kA. This would lower the voltage drop across the cable but it would require the power plant to raise its generated “line voltage” to 158 kv relative to ground to maintain the same power output (P=VI= 1000 MW). With a higher “line voltage” the cable can deliver the same amount of energy with less current, therefore minimizing the loss due to heating of the cable.

13. Thinking of the capacitor dielectric as a resistor, L=d and Acrosssection=Aplate, so R=1.2 x 1012 ohms.

a) The initial rate of discharge is dq/dt which equals the current flowing through the dielectric: I=V/R=8.3 x 10-11 amp.b) Using the definition of current and assuming steady discharge, I=Qcap/t t= Qcap/I= 59 sec.

14. a) You need to slice the cross-section block into horizontal

infinitesimal strips of thickness dy, then the current through

dr

each is J(5)dy. Since J increases linearly with “y” you can

determine the linear function for J and then you can integrate to get the answer, 2.5 A.

b) The electric field should be uniform and parallel to the current and can therefore be calculated from the expression E=J= J/. Then starting with R=V/I=EL/I R/L=E/I=10-4 ohm/m.

15.

You can start with P=I2R=(JA)2(L/A)=J2(AL) Power/Volume=J2.Since this final expression doesn’t depend on volume it should apply to any infinitesimal region of the material. This power changes electric potential energy to random heat energy and is referred to as Joule heating.

16.

a) This was done in class, check your notes, R=L/ab.

b) Since the current flows radially outward from the center, the elements of integration are thin cylinders of radius r, thickness dr, and height h, through which the current flows. R=(/2h)S.dr/r =ln(b/a)/2h.

17. From the previous chapter, C=Co[1-(-1)x/L] and Q=CV. Therefore Q depends on “x” and as the dielectric is pulled out, charge will flow out of the capacitor at a rate of dQ/dt=I in the wire. Using the chain rule dQ/dt= (dQ/dx)(dx/dt)= CoV(-1)v/L

Chapter 26: DC Circuits Hints & Answers-Updated last:3/28/08

1. Answer the following questions. Always explain your reasoning.

a) You should be able to show that in that way they have the minimum impact on the circuit they are trying to measure…b) Parallel…c) In series you are making a longer resistor, in parallel you are making a wider cross-section resistor…d) In series, I is the same and V is proportional to R (P=IP2PR). In parallel, V is the same and I is inversely proportional to R

(P=VP2/PR).e) Yes, it can be lower…f) Conservation of energy and of charge…g) To be able to compare the UrateU at which they charge …h) From eP-1P and from (1- eP-1P)…i) Probably will burn out…j) The light bulb probably won’t light…

2. Recall formula for E at surface of a cylindrical charge distribution: E=2k/r, where =q/L. We rearrange to get q=ErL/2k=1.67 nC. If there is a difference in current in and out of resistor, then q=(I)t. Solving for t=q/I=1.67 nC/1A=1.67 ms.

B

CD

A

B + - B

L V

l

B + - B

R RB2B

RB3

RB2 RB3

RB4

RB1

R

RB2RB1

1

-

1

-

RB1

RB3

RB2

RB4

1

-

3.

a) A=B<C=D. You should be able to conclude this from the fact the potentialacross A & B is less than for the other bulbs.You can also see this from the

fact that B & A have half the current of C & D.

b) C&D would be dimmer than before and B would be brighter than before, butall would have the same brightness since now they have the same potential.

c) C&D would dim less than before and B would be less bright. The reason is that in the second circuit the resistance is greater reducing the overall current. This causes the terminal voltage of the battery to increase and the potential across any elements is higher. The light bulbs all have the same brighness.

4. The theory of the voltage divider was done in class, check your notes.

a) VB1B=ERB1B/( RB1B+ RB2B); VB2B=ERB2B/( RB1B+ RB2B). This circuit is called a voltage divider because it divides the voltage of a source.

b) VBl B=E l/L

5. The limiting voltage on the resistors is V=(PR)P1/2P= 5 volts. We would need three resistors in series to reduce the voltage to 4v across each one. To reduce the overall resistance we put two more parallel branches of three resistors each.

6.

Illustration R’s in series R’s in parallel RBequivalentB Most powerful Ra) RB3 B & RB12B RB1 B & RB2B 3R/2 RB3B has more

voltage

b) RB3 B & RB2

later, RB1 B & RB234

RB4 B & RB23B 5R/3 RB1B has more voltage

c) NoneB NoneB Not applicable RB3B has more current

d) RB34 B & RB12B

(Actually all resistors have the same V and I so they are all in par & in series!)

RB1 B & RB2B

andRB3 B & RB4B

R All the same

RB1

RB3

RB2

RB4

1

-

RB1

RB3

RB2

RB4

1

- RB5

R vB + B R

12 k

RB, r B

CB1

RB2

RB1

CB2

1

- AB

R

e) RB1 B & R44B

andRB2 B & RB3

(Actually all resistors have the same V and I so they are all in par & in series!)

RB14 B & RB23B R All the same

f) RB34 B & RB12B;

(Actually all resistors have the same V and I so they are all in par & in series!)

RB1 B & RB2B

andRB3 B and RB4B

(RB5 has no current)

R RB1B -> RB4

are the same;RB5 has no current and no power

(g) When the R’s are equal, symmetry makes cases (d), (e), & (f) essentially the same circuit. With one different R, the symmetry is broken and the problem becomes more complicated. In (d) and (e) the same parallel and series combinations exist, but in (f) there is now a current through R5 and there are NO parallel or series combinations. You have to use Kirchhoff’s rules and, with 5 R’s with different currents, it could take 5 equations to solve!

7. The voltage across the 12-kresistor is 7.2 v without a voltmeter attached.

a) Voltmeter readings: 6.5v (9.2 % off); 7.1 v (1.9 % off), and 7.2 v (negligible error).b) The procedure is similar to above problem starting with a 3-k resistance

equivalent on the parallel branches. Voltmeter would further reduce the

equivalent resistance to 2.4-k …etc..

c) The ammeter would nearly “short” the 12-kresistor. Voltage acrossthe ammeter would be 0.176 v and the current through it 2.9 mA. If it weren’t

for the other resistors, the current would rise so high that it would overheat the

wires and could burn out the ammeter.

8..

a) Check the proof done in class.

b) E; 0; E.

c) 0; 0; (across the internal resistance”r” voltage is E)

9. Points A & B are at the same potential, then qB1B=qB2B. Also, there is a

single current passing through the resistors.

a) Potentials are the same across CB1B and RB1B, and across CB2B and RB2B. Setting up the

up the equations gives qB1B/ CB1B =IRB1B, and also q/ CB2B =IRB2.B

b) The series resistors make up a voltage divider: VBA B=VBBB= ERB2B/( RB1+B RB2B), with respect to graound.

R

R

2R

2R

1

- ABAB

R

R

R

2R

2R

CB

1

-

R

R

2R

2R

1

-

RB

R

R

2R

2R

1

- VB

R R

10. In the four circuits below R= 3 and E=12V. We assume ideal meters.

a) Simplify the circuit if you can and determine the potentials at A and B

with respect to ground. In the ammeter example A=B and you have two pairs

of parallel resistors in series. The capacitor and voltmeter case are basically

the same because no current flows between A & B.The hardest one to do is

the one with the resistor in the middle because you cannot simplify it and

you have to use K’s rules.However in this case you can simplify the problem

somewhat by using symmetry, the currents in the 2R resistors must be equal

and the currents in the side R resistors must be equal. The current in the center R must be the difference between these two and you have two unknowns in the Kirchhoff’s equations. Answers appear under diagram.

b) In the ammeter example A=B=6v, in all other A>B.

c) I haven’t done these yet. The ammeter, capacitor, & voltmeter examples are very similar to the previous cases. The one with the resistor in the middle is very hard to do because you need many equations, best to just set-up the equations and not bother solving them.

a) VBAB=8v; VBBB=4v VBAB=8v; VBBB=4v VBAB=6.86v; VBBB=5.14v

11.

a) Determine current in each case and use the power formulas. Lots of answers…b) Ditto. 13.8 W, 1.38 W, 0.138 W in the batteries and 1.38 W in the external resistor.c) This one will require K’s rules as each battery supplies a different fraction of the total current through the external resistor:

0.0154 W, 0.154W, 1.54 W in the batteries and 18.9 W in the external resistor.

12.

a) P=IP2PR= P2PR/(r+R)P2P

R

B1 + - B

1

- R

B2 + - B

1

-

R

RB2

RB1 RB4

1

-

1

-

b) Take the derivative of the expression above and set to zero to maximize: dP/dR=0 r=R. This means that when the internal and external resistances are equal the source is delivering the maximum amount of power to the external resistor. This however, also means that have the power is being wasted inside the battery itself.

c) When R>r more power in delivered to the external resistance than to the internal one, but there is less overall power generated. When R<r, more power is generated by the source but most of it is wasted inside the battery.

13.

a) The ammeter shors the resistor IB4B=0 and IBAB= 3A.

b) IB4B=1A and VBvB= 4 v.

c) You have two resistors in parallel IB4B=0.75A.

d) The capacitor is at first a short then it is an infinite resistance: IB4B=01A

e) Initially the potential across the resistor will be 12 v but eventually the capacitor will fill and become an infinite resistance: IB4B=3A1A.

f) IB4B=3A

14. The given conditions require that no current flow through EB3Beven when switch is

closed. This allows you to find the current through the other elements:

I= (1B+ 2B)/(RB1B+RB2B)

a) 3 B= 2B-IRB2B; or 3 B= - 1B+IRB1. b) No, there no potential across the switch whether it is closed or open in this

Situation. Note: Usually there is a potential difference across switches which is why they turn things on or off, but in this Uspecial caseU the switch serves no function, because there is no potential across those two points.

c) You would end with the same result as above because current would not flow through that branch.c) B3B.

15. In the illustrated circuit, take 1B=6.0V, B2B=1.5V, and 3B=4.5V. Also

RB1B=270, RB2B=150 RB3B=560and RB4B=820

a) IB3B=4.77 mA and is moving downward through EB2B.

b) With the same equations as used to solve previous part, state that IB3B>0 going up.

Working through the equation gives, B2B >5.49v.

15.5. (a) 0.113 ohm; (b) 6592 ohms

16.Measuremenst are rough: a) the source voltage=about 6.5v b) time constant=about 1 ms , and c) the capacitance =about 2 F.

RB1

CB2CB1

RB1

R

1

- CB1

RB3B

C2B

RB1

R

1

- CB

17. a) This was done in class: t/=ln100= 4.6

b) Starting with the expression for V= (1-eP-t/P) (dV/dt)= (eP-t/P/). At t=0, (dV/dt)= /= E /RC, and if this rate didn’t change /RC=V/t t =RC.

18. A 2.0F capacitor is charged to 150 V. It is then connected to an uncharged 1.0F capacitor through a 4.0 k resistor by closing the switch. Find the total energy in the resistor as

the circuit comes to equilibrium. You can do this in two different ways.

a) Recall that UBCB=VP2PC/2. Determine the initial U, then determine the final V and

final U, the difference must be the energy lost to heat at the resistor. U= 7.5 mJ.

b) The power in the resistor is P=iP2PR. Here the current decays exponentially according to the formula:

i=IBoB (eP-t/P). The heat generated can be determined by integrating the power from 0 to infinity: Heat=SPdt = S IBoPB2P(eP-

2t/P)dt= IBoPB2PR/2. You can show that this is the same as the answer to (a) once you establish that IBoB =VBoB/R and that the

time constant here is = RCB1BCB2B/( CB1B+CB2B) since the capacitors behave like a series combination while the current flows.

To derive the current formula start with loop rule around the circuit : q1/C1 – iR- q1/C1 = 0. You can turn this into a differential equation with respect to “i” by taking the derivative of this loop equation and using the fact that i= -dq1/dt= + dq2/dt.

19. a) Will be done in class. The general method is to integrate the power expression for the battery over time from t=0 to t=infinity ∫ idt). Then do the same thing for the resistor (∫ iP2PRdt) and compare the results. The work done by the battery will always be equally divided between the energy stored in the capacitor and the heat generated in the resistor.

20. Here the plate separation “d” is also the length L of the “resistor” created by the insulator.

a) RC= (d/A) (BBA/d) = BB.b) T=RCln2=1.6x10P5P sec= 1.8 days.c) The previous assignment had a similar problem in which it was assumed that the rate of the “leak” didn’t change. That

assumption was incorrect and the correct solution is that it would take an infinite amount of time to completely discharge the capacitor.

21. In the circuit illustrated the switch is initially open and both capacitors are initially uncharged. All the resistors have the same value R.

a) CB1 B shorts RB2B so the current through it is zero.b) Initially RB3B is shorted too but as the voltage in CB1B increases,Current the current through it increases. But this current eventually

Fills CB2B and the current drops to zero.

22. This will be discussed in class:

a) Initally the capacitor “shorts” RB2 Band the current passes only through

RB1 B(IBoB= /RB1B). After a “long time” the capacitor will be nearly

completely full and the current will pass through RB2B [IBooB= /(RB1B+ RB2B)].

b) IB1B goes from ( /RB1B) to /(RB1B+ RB2B); IB2B goes from 0 to /(RB1B+ RB2B);

IBcapB goes from ( /RB1B) to 0; all functions are exponentials.

c) Point rule: IB1B =IB2B +IBcap B; loop rules: - IB1BRB1 B- IB2BRB2B =0 ; - IB1BRB1 B– q/C=0 ; and IBcapB =dq/dt. These equations can be manipulated to generate one differential equation with one variable and the solution gives the given time constant.

d) This version is actually a lot easier that the above problem. The two parallel branches are essentially independent of each other. The current in RB2B will remain constant while the current in RB1B decreases exponentially over time as the capacitor fills up.