Lecture 4 Electric Potential/Energy Chp. 25•Cartoon - There is an electric energy associated with the position of a charge.•Opening Demo - •Warm-up problem •Physlet •Topics

•Electric potential energy and electric potential•Equipotential Surface•Calculation of potential from field•Potential from a point charge•Potential due to a group of point charges, electric dipole•Potential due to continuous charged distributions•Calculating the filed from the potential•Electric potential energy from a system of point charge•Potential of a charged isolated conductor

•Demos•teflon and silk•Charge Tester, non-spherical conductor, compare charge density at Radii•Van de Graaff generator with pointed objects

Electric potential• The electric force is mathematically the same as gravity so it too must be a

conservative force. We will find it useful to define a potential energy as is the case for gravity. Recall that the change in the potential energy in moving from one point a to point b is the negative of the work done by the electric force.

U = Ub -Ua = - Work done by the electric force =

• Since F=qE, U = and

• Electric Potential difference = Potential energy change/ unit charge

V= = U/q

• V = Vb -Va = - E.ds (independent of path ds)

⋅−b

a

dsEq

⋅−b

a

dsF

V = Vf -Vi = - E.ds (independent of path ds)

Therefore, electric force is a conservative force

U = Ub -Ua = - Work done by the electric force =

€

− F ⋅dsi

f

∫

V= = U/q

V = Vf -Vi = - E.ds (independent of path ds)

•The SI units of V are Joules/ Coulomb or Volt. E is N/C or V/m.

•The potential difference is the negative of the work done per unit charge by an electric field on a positive unit charge when it moves from point a to point b.

•We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at infinity for a point charge.

Examples• What is the electric potential difference for a positive charge moving in an

uniform electric field?

E

Ed

a bx direction

)( ab

b

a

b

a

xxEdxEdsEV −−=−=⋅−= d

V = -Ed

U = q V

U = -qEd

Examples• In a 9 volt battery, typically used in IC circuits, the positive terminal has a

potential 9 v higher than the negative terminal. If one micro-Coulomb of positive charge flows through an external circuit from the positive to negative terminal, how much has its potential energy been changed?

q

90−=−=

= ab VVqUV

qU 9−=

= -9V x 10-6 C

U = -9V x 10-6 Joules

U = -9 microJoules

Potential energy is lower by 9 microJoules

ExamplesA proton is placed in an electric field of E=105 V/m and released. After going 10

cm, what is its speed?

Use conservation of energy.

a b+

E = 105 V/m

d = 10 cm

V = Vb – Va = -Ed

U = q V

U + K = 0

K = -U

K = (1/2)mv2

(1/2)mv2 = -q V = +qEd

mqEdv 2

=

kgmC

v mV

23

519

1067.1110106.12

−

−

×⋅⋅×⋅

=

smv 6104.1 ×=

y 2

1

p r3 3

Electric potential due to point charges• Electric field for a point charge is E = (kq/r2). The change in electric potential a distance dr

away is dV = - E.dr.

• Integrate to get V = - kq/r2 dr = kq/r + constant.

• We choose the constant so that V=0 at r = .

• Then we have V= kq/r for point charge.

• V is a scalar, not a vector

• V is positive for positive charges, negative for negative charges.

• r is always positive.

• For many point charges, the potential at a point in space is the sum V= kqi/ri.

r1 r2

x

⎟⎠⎞⎜⎝

⎛ ++=3

3

2

2

1

1

rq

rq

rqkV

€

V f −Vi = − E ⋅dsi

f

∫

€

V f −Vi = − E ⋅drR

∞

∫ = −kq 1r2∫ dr = kq1

r R

∞

2rkqE =

RkqV =

041πε

=k

eqn 25-26

Replace R with r

rqV

041πε

=

€

0 −Vi = −k qR

Electric potential for a positive point charge

€

V (r) = kq /r

r = x 2 + y 2

Hydrogen atom. •What is the electric potential at a distance of 0.529 A from the proton?

• V = kq/r = 8.99*1010 N m2//C2 *1.6*10-19 C/0.529*10-10m

• V = 27. 2 J/C = 27. 2 Volts

Electric potential due to a positive point charge

+

-r r = 0.529 A

Ways of Finding V• Direct integration. Since V is a scalar, it is easier to evaluate V than E. • Find V on the axis of a ring of total charge Q. Use the formula for a point charge,

but replace q with elemental charge dq and integrate.

V = kdq/r = kdq/(z2+R2)1/2 = k/(z2+R2)1/2 dq = k/(z2+R2)1/2 Q

Point charge V = kq/r.For an element of charge dq, dV = kdq/r.r is a constant as we integrate.

This is simpler than finding E because V is not a vector.

V = kq/r

More Ways of finding V and E

•Use Gauss’ Law to find E, then use V= - E.ds to get V–Suppose Ey = 1000 V/m. What is V? V = -1000 y

•More generally, If we know V, how do we find E? dV= - E. ds –ds = i dx +j dy+k dz and dV = - Exdx - Eydy - Ezdz

Ex = - dV/dx,

Ey = - dV/dy,

Ez = - dV/dz.–So the x component of E is the derivative of V with respect to x, etc.–If Ex = 0, then V = constant in that direction. Then lines or surfaces –on which V remains constant are called equipotential lines or surfaces.–See examples

(Uniform field)

V = Ed

Equipotential Surfaces• Example: For a point charge, find the equipotential surfaces. First

draw the field lines, then find a surface perpendicular to these lines. See slide.

• What is the equipotential surface for a uniform field in the y direction? See slide.

• What is the obvious equipotential surface and equipotential volume for an arbitrary shaped charged conductor?

• See physlet 9.3.2 Which equipotential surfaces fit the field lines?

a) Uniform E field

E = Ex , Ey = 0 , Ez = 0

Ex = dv/dx

V = Ex d

V = constant in y and z directions

b) Point charge

(concentric shells)

c) Two point charges

(ellipsoidal concentric shells)

How does a conductor shield the interior from an exterior electric field?

•Start out with a uniform electric fieldwith no excess charge on conductor.Electrons on surface of conductor adjustso that:

1. E=0 inside conductor2. Electric field lines are perpendicularto the surface. Suppose they weren’t?3. Does E = ε just outside the conductor

4. Is uniform over the surface?5. Is the surface an equipotential?

6. If the surface had an excess charge, how would your answers change?

Dielectric Breakdown: Application of Gauss’s Law• If the electric field in a gas exceeds a certain value, the gas breaks down

and you get a spark or lightning bolt if the gas is air. In dry air at STP, you get a spark when E = 3*106 V/m. To examine this we model the shape of a conductor with two different spheres at each end:

21

+

+

+

Radius R

V = constant on surface of conductor

+

+

+

+

+

+

+

•The surface is at the same potential everywhere, but charge density and electric fields are different. For a sphere, V= q/(4π ε0 r) and q = 4πr2, then V = (/ ε0 )*r. Since E = / ε0 near the surface of the conductor, we get V=E*r. Since V is a constant, E must vary as 1/r and as 1/r. Hence, for surfaces where the radius is smaller, the electric field and charge

will be larger. This explains why:

21

+

+

+

Radius R

V = constant on surface of conductor

+

+

+

+

+

+

+

This explains why:• Sharp points on conductors have the highest electric fields and

cause corona discharge or sparks.

• Pick up the most charge with charge tester from the pointy regions of the non-spherical conductor.

• Use non-spherical metal conductor charged with teflon rod. Show variation of charge across surface with charge tester.

Van de Graaff+ + + +

Cloud

- - - -

Show lightning rod demo with Van de Graaff

21

+

+

+

Radius R

V = constant on surface of conductor

+

+

+

+

+

+

+

A metal slab is put in a uniform electric field of 106 N/C with the field perpendicular to both surfaces.

– Draw the appropriate model for the problem.– Show how the charges are distributed on the

conductor.– Draw the appropriate pill boxes.– What is the charge density on each face of

the slab?– Apply Gauss’s Law. E.dA = qin/ε0

- - - - -

.Right side of slab 0*A + E*A = A/ε0,

E = /ε0, = 106 N/C *10-11 C2/Nm2 = +10-5 N/m2

(note that charges arranges themselves so that field inside is 0)

++++

GaussianPill BoxSlab of metal

In a uniform Electricfield

E = 106 N/C

Slab of metal In a uniform Electric field

Evaluate E.dA = qin/ε0

• Left side of slab - E*A + 0*A = A/ε0,

E = - /ε0, = - 106 N/C *10-11 C2/Nm2 = -10-5 N/m2

What is the electric potential of a uniformly charged circular disk?

Warm-up set 4 1. HRW6 25.TB.37. [119743] The equipotential surfaces associated with an isolated point charge are: concentric cylinders with the charge on the axis vertical planes horizontal planes radially outward from the charge concentric spheres centered at the charge

2. HRW6 25.TB.17. [119723] Two large parallel conducting plates are separated by a distance d, placed in a vacuum, and connected to a source of potential difference V. An oxygen ion, with charge 2e, starts from rest on the surface of one plate and accelerates to the other. If e denotes the magnitude of the electron charge, the final kinetic energy of this ion is: 2eV eVd Vd / e eV / d eV / 2

3. HRW6 25.TB.10. [119716] During a lightning discharge, 30 C of charge move through a potential difference of 1.0 x 108 V in 2.0 x 10-2 s. The energy released by this lightning bolt is: 1.5 x 1011 J 1500 J 3.0 x 109 J 3.3 x 106 J 6.0 x 107 J