Lecture 6-1Today

• Electric “Shielding”

• Electric Potential

Lecture 6-2 Electrostatic “Shielding” I

If you move charge q in the cavity,

the exterior electric fields and the

extreior charge distribution are not

affected.

Conducting shell electrostatically

“shields” its exterior from

changes on the inside.

• As long as the E contribution due

to all the interior charges (both in

void area and on interior surface),

measured at the outer surface

remains zero, then moving the +q

charge does not affect the exterior

surface charge distribution.

Truth:

Superposition,

NOT shielding!

q

Lecture 6-3Electrostatic Shielding II

Conducting shell electrostatically

“shields” its interior from changes

in the exterior, and vice versa.

If you now add charge Q’ to the conductor

and/or Q’’ on the outside of the conductor,

the interior electric fields do not change.

Add Q’Q’’ Q’Q’’

Truth:

Superposition,

NOT shielding!

Like active noise

cancellation

+++

+

+

+

Lecture 6-4

Electric Potential Energy of a Charge in Electric Field

• Coulomb force is conservative

=> Work done by the Coulomb

force is path independent.

• Can associate potential energy

to charge q0 at any point r in

space. ( )U r

It’s energy! A scalar measured in J (Joules)

d l

0dW q E dl= Work done by E field

0dU dW q E dl= − = − Potential energy change

of the charge q0

Lecture 6-5Electric Potential Energy of a Charge (continued)

0dW q E dl=

0

( ) ( )

r

i

U U r U i

q E dl

= −

= −

0dU dW q E dl= − = −

i is “the” reference point.

Choice of reference point

(or point of zero potential

energy) is arbitrary.

0

d l

i is often chosen to be

infinitely far ( )

Lecture 6-6

Gravitational vs Electrostatic Potential Energy

( ) ( )

b

a

U U b U a

dF l

= −

= −

a

b

qEmgGravity Coulomb

Work done by<gravity/Coulomb> force is the

decrease in potential energy.

mg l− qE l−

(if g, E uniform)

Lecture 6-7

Potential Energy in the Field due to a Point Charge q

0

0

2

0

2

0 0

( )P

P

r

r

U r q E dl

q qk r dl

l

q qk dl

l

q q q qk k

l r

→

→

= −

= −

= −

= =

From ∞

This is also called the potential energy of

the two-charge configuration of q and q0.

What is the work required to

separate the two charges to ?

Independent of path since

static E is conservative.

Lecture 6-8

Potential Energy of a Multiple-Charge Configuration

(b)

(a)1 2 /kq q d

1 31 32 2

2

q q qq qk k k

dd

q

d++

(c)

2 3

1 3 3 41 2 2 4

1 4

2 2

q q q qq q q qk k k k

d d

q q q qk k

d

d d

d

+

+

+

+

2 d

Lecture 6-9 Electric Potential and Electric Field

1

1

1

r

q

iq

UV E dr

= = −

Normalize by the

probe charge q1:

1 int 1int

f f

q

i i

U W F d qr E dr = − = − • = −

Potential energy change of a probe charge q1 in the

electric field E created by all the other (source) charges

[Electric potential] = [energy]/[charge]

SI units: J/C = V (volts)

Potential energy difference when 1 C of charge is

moved between points of potential difference 1 V1 J =

Scalar!

independent

of q1

Lecture 6-10 Electric Potential

• So U(r)/q0 is independent of q0, allowing us to introduce

electric potential V independent of q0.

• [Electric potential] = [energy]/[charge]

SI units: J/C = V (volts)

• U(r) of a test charge q0 in electric field generated by

other source charges is proportional to q0 .

0

( )( )

U rV r

q

taking the same

reference point

Potential energy difference when 1 C of charge is

moved between points of potential difference 1 V1 J =

Scalar!

1

1

0

r

q

iq

UV E dr

= = −

Lecture 6-11

Potential at P due to a point charge q

0

0

( )( )

qU rV r

q

qk

r

=

=

From ∞

Lecture 6-12

Electric Potential Energy and Electric Potential

positive

chargeHigh U

(potential

energy)

Low U

negative

charge

High U

Low UHigh V

(potential)

Low V

Electric field direction

High V

Low V

Electric field direction

Lecture 6-13

Volt (potential) and Electron Volt (energy)

• V=U/q is measured in volts => 1 V (volt) = 1 J / 1 C

J N mV E m V

C C

N VE

C m

= = = =

= =

19

1 1 1

1 | | 1 1.602 10 1

J C V

eV e V C V−

=

• V depends on an arbitrary choice of the reference point.

• V is independent of a test charge with which to measure it.

(electron volt)

Lecture 6-15

Potential due to two (source) charges on their axis

1 2( )| | | |

q qV x k k

x x a= +

−

1 2 0q q=

Lecture 6-16

Potential due to Multiple Source Charges: Example

1 2 3 4

( )

/ 2

V P

q q q qk

d

=

+ + +

Dotted line is an equipotential when

q1=12nC, q2= -24nC, q3=31nC, q4=17nC

E from V

We can obtain the electric field E from the potential V by

inverting the integral that computes V from E:

( ) ( )

rr

x y zV r E dl E dx E dy E dz= − = − + +

If r is in x-direction, e.g., then dy=dz=0 and Ex is

negative of the rate of change of V in x-direction.

V(x)

x

0xE =