1/29/2020 2

Calculate Electric Field from the Potential

Electric field always points in the direction of steepest descent of V (steepest slope) and its magnitude is the slope.

Potential from a Negative

Point ChargePotential from a Positve

Point Charge

-V(r )

y

x

V(r )

y

x

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Calculating the Electric Field from the Potential Field

If we can get the potential by integrating the electric field:

We should be able to get the electric field by differentiating the potential.

b

a

ba ldEV

VE

In Cartesian coordinates:

dx

VEx

dy

VEy

dz

VEz

In the direction of

steepest descent

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Calculating the Electric Field from the Potential

Field

ˆˆ ˆ

points in the direction in which

the potential decreases most rapidly.

From this we see that

, , and .x y z

V V VE V i j k

x y z

E

V V VE E E

x y z

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Potential from a Continuous Charge Distribution

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total chargeQ

small piecesof charge

dq

Line of charge: = charge per unit length [C/m]

dq = dx

Surface of charge: = charge per unit area [C/m2]

dq = dA

Volume of Charge: = charge per unit volume [C/m3]

dq = dV

Charge Densities

dq rdrddz

dq r2dr sindd

Cylinder:

Sphere:

dq rddz

dq r2 sindd

Cylinder:

Sphere:

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Calculate Potential on the central axis of a

charged ring

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Calculate Potential on the central axis of a

charged disk

dq A 2a da

V kdq

r

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Example: Calculating the Electric Field from the

Potential Field

What is the electric field at any point on the central

axis of a uniformly charged disk given the potential?

Given: V

20

( z2 R2 z)

VE

Ex

Ey

Ez

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Calculate Potential on the central axis of a

charged disk (another way)

V E dl

Ez

2o

1z

z2 R2

From Lecture 3:

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Calculate Potential due to an infinite sheet

++++++++++++++++

ˆˆ ˆ ˆ2 ( )

2 2

dV E dl

dV k i dx i dx j dx k

V k dx k x C

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E due to an infinite line charge

Corona discharge around a high voltage power line,

which roughly indicates the electric field lines.

August 26, 2014: Four Kentucky Firefighters Electrocuted

When Participating in ALS Ice Bucket Challenge

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V due to an infinite line charge

R

Lline charge with

charge density

Gaussian

cylinder

Remember Gauss’s Law

s

o

insiden

s

QdAEAdE

EndcapsbarrelRs

n dAEdAEdAE0

o

RBarrelR

LRLEAE

)2(

LQinside

R

kE

RE

R

o

R

2

or2

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V due to an infinite line charge (continued)

dV E dl ERˆ R dl ERdR

R

RkV

R

RR

R

Rk

R

dRk

dREVV

ref

ref

refref

ref

P

R

R

R

RRrefP

P

ref

P

ref

ln2

and 0

.ln and 0ln

because or 0

chooset can'you :Problem

ln2

2

Where we define V = 0 at R = Rref

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Equipotentials

Definition: locus of points with the same potential.

•General Property: The electric field is always

perpendicular to an equipotential surface.

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Equipotentials: Examples

infinite positive

charge sheet

electric dipolePoint charge

V(x) 2kx Vo

V(r) kq

r

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+++++++++++++++

–––––––––––––

Locally

Gauss:

at electrostatic

equilibrium

E

0

E|| 0

in electrostatic equilibrium

all of this metal is an equipotential;

i.e., it is all at the same voltage

Equipotential Lines on a Metal Surface

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Potential inside & outside a conducting sphere

The electric field is zero inside a conductor.

The electric potential is constant inside a conductor.

Vref 0

at r .

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The graph gives the electric potential V as a function of

x. Rank the five regions according to the magnitude of

the x-component of the electric field within them,

greatest first.

Quiz 3

1 2 3 4 5

V

x

(A) 4 > 2 > 3 > 1 > 5

(B) 3 > 1 > 5 > 2 = 4

(C) 2 > 4 > 1 = 3 = 5

(D) 1 = 2 > 4 = 3 > 5

(E) 1 = 3 = 5 > 2 > 4

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electric field

linesV1 100 V

V2 80 V

V3 60 V

V4 40 V

Work and Equipotential Lines

Which of the following

statements is true about the

work done by the electric

field in moving a positve

charge along the paths?

(A) III = IV > 0, I = II = 0

(B) I = II = 0, III > IV

(C) III = IV < 0, I = II = 0

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Summary

• If you know the functional behavior of the potential V at any point, you can calculate the electric field.

• The electric potential for a continuous charge distribution can be calculated by breaking the distribution into tiny pieces of dq and then integrating over the whole distribution.

• Finally no work needs to be done if you move a charge on an equipotential, since it would be moving perpendicular to the electric field.

• The charge concentrates on a conductor on surfaces with smallest radius of curvature.