electrostatics slow learner

8/13/2019 Electrostatics Slow Learner

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STUDY MATERIAL FOR SLOW LEARNERS

1. How many electrons will have a total charge of 1 coulomb?

Given, q = 1 C, n = ? From q = n e

n = q = 1 = 6.25 1018e 1.6 10-192. Find the electric field intensity due to a charge of 5 10-8C at a point 50 cm from it in

vaccum.

Given, Q = 5 10-8C r = 0.5 mE = 1 Q

4 o r2 = 9 1095 10-8(0.5)2

3. an electric charge of 2 c experience an electric force of 3.2 10-3N when kept at a point invaccum. What is the electric field intensity at that point?

Given, q = 2 10-6CF = 3.2 10-3N

E = F = 3.2 10-3q 2 10-6 =1600 N / C

4. An electric dipole consists of two charges of magnitude 1 c. They are placed 3 cm apart in auniform electric field of 100 N C-1 acting at right angles to the axis of dipole. Calculate(i) dipole moment of electric dipole ii)electric force acting on either chargeiii) Torque acting on dipole

Given, q = + 10-6C and106C; 2a = 3 10-2m; = 90o E = 100 N/C(i) Dipole moment , p = q 2a = 10 -63 10-2= 3 10-8cm(ii) Force acting on each charge, F = q E = 1-6100 = 10-4N(iii) Torque, = P E Sin = 3 10-8100 Sin 90o= 3 10-6Nm5. An electric dipole is held at an angle of 30owith respect to a uniform electric filed of 2 104

NC-1experiencing a torque of 18 10-25Nm. Calculate the dipole moment of the dipoleGiven : = 30o ; E = 2 10-4N/C ; = 18 10-26N m

As = P E Sin P = / E Sin

= 18 10-262 10-4Sin30o = 1.8 10-29cm

q = 5 10-6 ; E = 50 V/m; 2a = 2 Cm = 2 10-2m(a) Dipolement, p = q 2a = 5 10-62 10-2 = 10-7cm(b) Initial torque acting on the dipole, = P E Sin = 10-750 = 2.5 10-6N m(c) Initial potential energy of dipole = - P E Cos = 10-750 3/2 = 4.33 10-6J(d) Potential energy in stable equillbrium, W = - P E = - 10-750 = - 5 10-6J

Potential energy in unstable equilibrium, W = P E = 50 10-7= 5 10-6J(e) Change in potential energy when it turns from its initial position and aligns along E

Sur = W2W1= - P E( - PE Cos )= - PE + PE Cos = -5 10-6+ 4.33 10-6 = 0.67 10-6J

(f) Work done in turning it from its stable to unstable position

W = W2W1= P E( - P E ) = 2 PE = 10 10-6J6. Electric potential at a point is 200 Volt. How much work will be down in bringing an particle

from infinity to that point (Charge on an - particle is 3.2 10-19C)Given, V = 200 V , q = 3.2 10-19 C; From, V = W/q

W = v q = 3.2 10-19200 = 6.4 10-17J7. Find the potential at a point 30 cm from a pint charge of 2 10-8C in vacuum

Given q = 2 10-8C, r = 0.3 mV = 1 q

4 o r= q 1092 10-8


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3 = 600 Volt9. An electric field at a point due to a point charge is 20 N / C and the electric potential at thatpoint is 10 J / C. Calculate the distance of the point from the charge and also the magnitude of8he charge.

Let r be the distance of pint from the source charge q.

1 q = 20 ------4 o r2

1 q = 10 --------

4 o rDividing equation by equation we get

R = 10 / 20 = 0.5 m

Putting r = 0.5 m in equation

9 109q = 100.5

q = 0.5 10 = 5.56 10-10C9 107

9. A Cube of side b has a charge q at each of its eight vertices. Determine the potential and

electric field due to these array at the centre of the cube.

Distance of centre of the cube from each vertex x = b2

3

Potential at the centre due to all the 8 charges, V = 8 1 q4 o x

= 8 q 24 o b 3

V = 4q / o b 310. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of80.0 C. m-2. Find (a) The charge on the sphere and (b) total electric flux leaving the surface ofsphere.

Given, r = 2.4 / 2 = 1.2 m; = 80 10-6cm-2(i) From q = A = r24 = 80 10-64/7 22 (1.2)2= 1.45 10-3C(ii) Applying Gausss Theorem to a spherical Gaussian surface enclosing the charged sphere

Electric flux through the Gaussian Surface i.e., Surface of Sphere

= q / o= 1.45 10-3

8.85 10-12 = 1.6 108N / C m 211. What is the area of the plate of the plate of a 2 F parallel plate capacitor with plate

separation of 0.5 Cm? Why do ordinary capacitor have capacitance of order of microfarads?Given, C = 2 F; D = 0.5 10-2 m ; From C = A o/ d

A = C d / o= 2 10-20.5

8.85 10-12 = 1130 106 m2 A = 1130 Km212. A 4 F capacitor is connected in parallel with a 8 F capacitor. The combination is chargedat 300 volt. Calculate (i) the total charge on the combination (ii) the total energy stored in thecombination

Given, C1= 4 F = 4 10-6F ; C2= 8 10-6F ; V = 300 V(i) Net capacitance of the parallel combination, C = C1+ C2= 4 + 8 = 12 F

Total charge on the combination, q = C V = 12 10-6300 = 3.6 10-3C(ii)Total energy stored in combination , V = C V2

= 12 10-69 104= 0.54 J


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13. The plates of a parallel plate capacitor have an area of 90 cm2each and are separated by 2.5mm. (a) Find the capacitance of capacitor (b) If the capacitor is charged by connecting it to a 400V supply, how much energy is stored by the capacitor? (c) Calculate the energy stored per unitvolume of the capacitor

Given A = 90 10-4m2 ; D = 2.5 10-3m(a) C = A o/ d

= 90 10-48.85 10-122.5 10-3 = 31.86 10-12F(b) Energy stored in the capacitor U = CV2 = 31.86 16 104 = 2.5 10-6J

(c) Energy stored per unit volume = U / Ad

= 2.55 10-69 10-32.55 10-3 = 0. 113 Jm-3

14. Two parallel plate capacitors of 20 F and 30 F are charged to 30 V and 20 V respectively.If the plates of these capacitors with same type of charge are connected together. Find(a) The common potential of the capacitors(b) Charges on the capacitor at common potential(c) Loss of energy in the processGiven, C1= 20 10-6F; C2= 30 10-6F; V1= 30 V ; V2= 20 V(a) Common potential,

V = C1V1+ C2V2

C1+ C2= 20 10-630 + 30 10-620

50 10-6 = 24 V(b) Let the charges on the capacitors when they had attained common Potential be q1land q21

q1l= C1V = 20 10-624 = 4.8 10-4C and q2l= C2V = 30 10-624 = 7.2 10-4C(c) Energy lost

= C1C2(V1V2)22(C1+C2)

= 20 10-630 10-6( 3010 )22 ( 20 + 30 ) 10-6 = 6 10-4J

15. An infinite line charge producer a field of 9 104N / C at a distance of 2 cm. Calculate thecharge density. Given E = 9 104N / C , r = 2 cm = 2 10-2m

From E = 2 or = E 2 or

= 9 104 1 2 10-218 109

2 o= 1/ 18 109 = 10-7C Or 0.1 C


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ELECTROSTATICSQUESTIONS FOR PRACTICE

1. Coulombs lawin vector form: 11 q1q 2

F = r

4 r22. In a medium the force of attraction between two point charges, distance d apart is F.What distance apart should these be kept in the same medium so that the force betweenthem become (a) 5 F (b) F ? 5

Ans:- F 1d2

(a) For the force to become 5 F, the separation of must become d root 5

(b) For the force to become F / 5, the separation d must become d / 5

3. Sketch the electric lines of force for (a) 5 C (b) 10 C

Less lines of force More lines of Force

4. Always the neutral point lie closer to the smaller charge when we draw lines of forcebetween similar charges.

5.

What is the relation between EA, EB& EC.

EA > EB > EC

6. If a point charge +q, is taken first from A to C and then from C to B of a circle drawn withanother pointcharge + q as centre, then along which path more work will be alone?

WD will be same, as the distance is same.7. An uncharged insulated conductor A is brought near a charged insulated conductor B.What happens to charge and potential of B?

Potential decreases but charge remains small as both are insulated.

8. A and B are two conducting spheres of same radius, A being solid and B hollow. Bothare charged to the same potential. What will be the relation between the charges on thetwo spheres?

As charges resides on the surface of the conductors, both the spheres contains

same amount of charges.9. A conductor with a cavity in it is given a charge Q. What will be the total charge on thesurface of this conductor if another conductor carrying charge q is placed in the cavitywithout touching the cavity>

The charges get added up i.e., q + Q is the charge.


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10. A capacitor of capacitance c has distance between plate is d. A very thin wire mesh isplaced as shown in figure. Calculate new capacitance.

Cp = C1C2 = 2 C . 2 C = 4 C2 = CC1+C2 2C + 2C 4 C

Cp = C, Remains the same2 MARKS & 3 MARKS

1. A point charge q is placed at O as shown. Is VAVBpositive, negative or zero, if q is a(i) positive (ii) negative charge?

O A Bq

(i) When q is positive, VAVBis positive.(ii) When q is negative, VAVBis negative.

2. An electric dipole is free to move in a VEF. Explain its motion when it is placed (1)parallel to the field and (ii) perpendicular to the field

(1) F1 F2 E

+q - qEither it may be stable or unstable equilibrium. Also, there is no torque or

translatory force.+q F1

(2)

EF2

- q

The dipole will experience a torque. As EF is uniform, there is no translatory force.

3. Two capacitors of equal capacitance, when connected in series, have a net capacitance ofC, and when connected in parallel have a capacitance of C2. What will be the value of C1/C2.

(i) Series 1 = 1 + 1 C1 = CC1 C C 2

(ii) Parallel C2= C + C C2= 2 C

C1 = C/2 = 1C2 2C 4

C1 : C = 1 : 4

4.

Find the resultant capacitance


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1 = 1 + 1 + 1CR C C C

This circuit is equal to three capacitors connected in parallel.

5. A spherical Gaussian surface encloses a charge of 17.7 X 10 8C. (i) Calculate theelectric flux passing through the Gaussian Surface. (ii) If the radius of the Gaussian

surface is double, how much flux would pass through the surface?

(i) = q = 17.7 X 10 -8 = 2 X 104 Nm2C -1

o 8.85 X 10 12(ii) As flux is not related to Gaussian surface there is no change in electric flux.

6. Two point charges q and 2q are kept at a distance d apart from each other in air. Athird charge Q is to be kept along the same line in such a way that the net force acting on qand 2q is zero. Calculate the position of charge Q in terms of q and d.

d

x1 x2q Q 2q

For the net force on q and 2q to be zero, Q must be between q and 2q and it must beof negative sign.

1 q . 2q = 1 q. Q = 1 Q. 2q

4 o d2 4 o x12 4o x22

x22 = 2x12

Substituting it in and simplifying

x1 = d and x2 = 2 d1 + 2 1 + 2

7. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to(1/2) QE, where Q is the charge on the capacitor and E is the magnitude of EF between theplates. Explain the origin of the factor .

Work done in increasing the separation between the plates by a distance dx againstthe force of attraction F between the plate is

dw = F dx ------------

Energy density, u = o E2

Increase in PE = u. A. dx = o E2 A dx

= (o E ) A E dx = AE dxWhere, E = , = Q -----------

o AFrom & , F dx = QE dx

F = QEThe factor denotes that plates are being moved against an average value E/2.


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8. Calculate capacitance of capacitor as show.

The equivalent circuit is

Using the general formula , C = o K AD

By applying series and parallel combination formula.

Resultant Capacitance = o A K1 + K2 K3D 2 K2+ K3

9. Derivation for Energy Density in a parallel plate capacitor.

Energy Density = Total Energy StoredVolume

= CV2

Ad

But, C = A o V = Edd

Therefore, u = A o E2d2

d

Ad

u = oE2

When the dielectric medium is kept between the plates, then

u = o r E210. Energy stored in series combination.

In series combination, the charge q is same on each capacitor.


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Therefore, Energy stored in the series combination of n capcitors of capacitances,C1, C2, Cn will be

W = q2 = q2 1 + 1 + . + 12C 2 C1 C2 Cn

W = q2 + q2 + . q22C1 2C2 2Cn

Therefore, W = W1 + W2 + + Wn

11. Energy stored in parallel combination. In parallel combination, the pd across eachcapacitor is same.

Therefore, Energy stored in parallel combination

W = Cp V2 = (C1 + C2 + . + Cn ) V2

W = C1V2 + C2V2 + . + Cn V2

W = W1 + W2 + W3 + .. + Wn

Total energy stored = Sum of the energies stored in individual capacitors.

12. Concept of common potential (Sharing of charges)

When two capacitors initially charged to different potential are connected across

each other, redistribution of charges takeplace. There is flow of charges, till they attain thecommon potential.

Initially, C1 Charged to potential V1 with Charge q1C2 charged to potential V2 with charge q2

q1 = C1 V1, q2 = C2 V2Let V2 > V1

Common Potential = V = Total charge on capacitorsTotal Capacitance

V = q1 + q2C1 + C2

V = C1V1 + C2 V2C1 + C2

Note: - New charges in C1& C2 areq1l= C1V & q2l = C2V13. Loss of Energy while sharing of charges.

When two charged capacitors are connected to each other, the charge flows fromone to other, till they attain common potential while sharing charges some amount ofenergy is lost in the form of heat.

Total energy of the system before connecting = W1= C1V12 + C2V22Total energy after attaining common potential, W2 = (C1+ C2) V2

= (C1+ C2) C1V1+ C2V2 2C1 + C2

= (C1V1+ C2V2)2

2 ( C1+ C2)


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As W is positive, There is always loss of energy while sharing of charges.

14. Relation between e and r

r = 1 +

e

15. The diagram shows the arrangement of plates of a parallel plate capacitor. Find theresultant capacitance of this arrangement.

C1 = o A/2 = o Ad 2d

C2 = o A/2 = o A2d 4d

C = C1+ C2 = o A + o A = 2 o A + o A = 3 o A2d 4d 4d 4d

C = 3 o A4d

16. What is the equivalent capacitance of the system of plates?

C1= C2= o Ad

C = C1+ C2 = 2 o Ad


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17. What is the equivalent capacitance of the system of plates?

C1= C2= C3 = o Ad

C = C1+ C2+ C3 = 3 o Ad

18. Capacitors P, Q and R have each a capacitance C. A battery can charge the capacitor Pto a potential difference V. If after charging P, the battery is disconnected from it and thecharged capacitor P is connected in following separate instances to Q and R (i) to Q inparallel and (ii) to R in series, then what will be potential difference between the plates of Pin the two instances.

Ans: When the capacitor P is charged to a potential difference V, then

Charge acquired by the capacitor q = CV

(i) When capacitor P is connected to Q in parallel when the capacitors P and Q of eachcapacitance C are connected in parallel

Total capacitance = C + C = 2C

Total charge on the two capacitor = q + 0 = q = CV

Therefore Pd across each capacitor or the capacitor P will become

Vl

= CV = V2C 2

(ii) When capacitor P is connected R in series, The circuit will not completed. So there is

no sharing of charges.

The Pd across P remains same.

electrostatics slow learner

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