Physics 202, Lecture 6Today’s Topics

Electric Potential (Ch. 23) Electric Potential For Various Charge Distributions

Point charges Continuous chargese.g Uniform Ring, Sphere, Shell

Equipotential surfaces, Electrostatic equilibrium of conductors

Expected from preview: EV relationship, equipotential lines, electrostatic

equilibrium of conductors… Electric potential for a charge distribution…

Review: Electric Potential Difference Electric Potential Energy: q In a Generic E. Field

Electric Potential Difference

VqdqUUUB

AAB Δ=•−=−=Δ ∫ sE

test sourcesystem energy

AB

B

AVVd

qUV −=•−=Δ≡Δ ∫ sE

source only

Electric Potential Difference For Single Point Charge (Lecture 5 Review)

VB-VA = ke(q/rB-q/rA)

A

B

rA

rB

q

Take V∞ =0

V = keq/r

Electric Potential ForContinuous Charge Distribution

For finite charge distribution, it is common to setV=0 at infinite.

If the charge distribution is known, V can be calculated simply by scalar integral.

dV=ke dq/r (V=0 @ ∞)

∫=rdqkV e

Example: Uniformly Charged Ring For a uniformly charged ring, show that the

potential along the central axis is

22 axQkV e

+=

∫

∫

∫

+=

+=

=

dqax

kax

dqkrdqkV

e

e

e

22

22

Solution

=Q

Uniformly Charged Ring: Electric Field

( )symmetry to due 0

cos

2/3223

2

=+

===

==

⊥

∫E

axxQk

rxQkdEE

rx

rdqkdEdE

eexx

ex θ

Find the electric field along the central axis.Approach 1: Superposition.

Approach 2: derivative of potential

( ) 2/322 axxQk

xVE e

x+

=∂∂−=

Example: Uniformly Charged Spherical Shell

For uniformly charged spherical shell.Again, use:

RrrQkE e

r >= 2

RrE <= 0

Tip: V is the same inside E=0 region

AB

B

AVVdV −=•−=Δ ∫ sE

R

Example: Uniformly Charged Sphere Show that for a uniformly charged

sphere, the electric potential is: R

Hint: It is more convenientto use:

since from Gauss’s law:

AB

B

AVVdV −=•−=Δ ∫ sE

RrrQkE e

r >= 2

RrRQrkE e

r <= 3

Exer

cise w

ith T

A

Coulomb’s Law: (lecture 2)

Gauss’s Law: (lecture 3,4)

Potential: (lecture 5,today)

E = k dq

r2r̂∫

EidA∫ =

qenclε0

V = k dqr∫

E = −∇V

Ex = − ∂V∂x ,Ey = − ∂V

∂y ,Ez = − ∂V∂z

Summary: Methods for Obtaining E, V

V = −

Eidl∫

Know all methods (and when to apply which)!

Conductors in Electrostatic EquilibriumWe have learned that:

Electric field inside conductor is zero (also zeroinside any empty cavity within the conductor)

All net charges reside on the surface Electric field on surface of conductor always

normal to the surface, magnitude σ/ε0

sharper edge larger field.

Since E=0 inside conductor, Potential is the same throughout the conductor:

Equipotential

Charge Distribution on Conductors:Field lines and Equipotential Surfaces

Review: Charge Distribution On Conductors (I)

|E| higher at smaller radius of curvature (more charge density)

|E| lower(less charge density)

High Voltage Electrostatic Generator:Van de Graaff Generator

104V

up to 3x106V(~50KV in today’s demo)

Example

A B

Two conductors are connected by a wire. How do the potentialsat the conductor surfaces compare?

a) VA > VB b) VA = VB c) VA < VB

What happens to the charge on conductor A after it isconnected to conductor B ?

a) QA increases b) QA decreases c) QA doesn’t change