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QUIZ 6 January 26, 2012 An electron moves a distance of 1.5 m through a region where the electric field E is constant and parallel to the displacement. The electron’s potential energy increases by 3.2×10-19 J. What is the magnitude of E ?

a)1.3N/Cb)5.0×10-19N/Cc)2.0N/Cd)1.0×10-19N/Ce)7.5×10-1N/C

e = 1.6×10-19 C k = 8.99×109 Nm2/C2 ε0 = 8.85×10-12 C2/(Nm2) 1/30/12 2

Chapter 24 •  Electrostatic Potential Energy of a system of fixed

point charges is equal to the work that must be done by an external agent to assemble the system, bringing each charge in from an infinite distance.

q1 Point 2 Point 1

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Example: Electrostatic Potential Energy

q1 Point 2

q2 Point 1

If q1 & q2 have the same sign, U is positive because positive work by an external agent must be done to push against their mutual repulsion.

If q1 & q2 have opposite signs, U is negative because negative work by an external agent must be done to work against their mutual attraction.

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Example: Three Point Charges

q1 Point 2 r1,2

q2 Point 1

r1,3 r2,3

Point 3

q3

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Electrostatic Potential Energy

•  We can conclude that the total work required to assemble the three charges is the electrostatic potential energy U of the system of three point charges:

3,2

23

3,1

13

2,1

1232 r

qkqrqkq

rqkqWWWU total ++=+==

The electrostatic potential energy of a system of point charges is the work needed

to bring the charges from an infinite separation to their final position

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Calculate Electric Field from the Potential Electric field always points in the direction of

steepest descent of V (steepest slope) and tis magnitude is the slope.

Potential from a Negative Point Charge

Potential 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

E = −

∇V = −

∂V∂x

i +∂V∂y

j +∂V∂z

k⎛⎝⎜

⎞⎠⎟

Ex = −∂V∂x

, Ey = −∂V∂y

, and Ez = −∂V∂z

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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?

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Potential due to a Group of Point Charges

1q

2q3q 4q

1r

2r3r

4r

X

V (r) = Vn (r) =14πεon=1

N

∑ qnrnn=1

N

∑

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Potential from a Continuous Charge Distribution

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total charge Q

small pieces of charge dq

Lineofcharge:= 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 = ρrdrdθdzdq = ρr2dr sinθdθdφ

Cylinder:

Sphere:

dq = σrdθdzdq = σr2 sinθdθdφ

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

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Calculate Potential on the central axis of a charged disk (another way)

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Calculate Potential due to an infinite sheet

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

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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.

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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 dipole Point charge

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

–––––––––––––

Locally

Gauss: at electrostatic equilibrium �

E⊥ = σε0

�

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

Common mistake: thinking that potential must be zero inside because electric field is zero inside. �

Vref = 0 at r = ∞.

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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.

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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 7 Jaunary 31, 2012

1 2 3 4 5

V

x

(A)  4 > 2 > 3 > 1 > 5

(B)  3 > 1 > 5 > 2 = 4

(C)  1 = 2 > 4 = 3 > 5

(D)  2 > 4 > 1 = 3 = 5

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Another Method: Electrostatic Potential Energy Leave the charges in place and add all the

electrostatic energies of each charge and divide it by 2.

where:

qi -- is the charge at point i

Vi -- is the potential at location i due to all of the other charges.

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Another Method: Electrostatic Potential Energy

r1,3 r2,3

Point 3

q3

q1 Point 2 r1,2

q2 Point 1

U =12q1

kq2r1,2

+kq3r1,3

⎛

⎝⎜⎞

⎠⎟+ q2

kq1r1,2

+kq3r2,3

⎛

⎝⎜⎞

⎠⎟+ q3

kq1r1,3

+kq2r2,3

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

U =122 kq1q2r1,2

+ 2 kq3q1r1,3

+ 2 kq2q3r2,3

⎡

⎣⎢

⎤

⎦⎥ =

kq1q2r1,2

+kq3q1r1,3

+kq2q3r2,3

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Summary: Electrostatic Potential Energy for a Collection of Point

1. Bringing charge by charge from infinity

2. Or using the equation: and trying not to forget the factor of ½ in front.

Note: The second way, could involve more calculations for point charges. However, it becomes very handy whenever you try to calculate the electrostatic potential energy of a continuous distribution of charge, the sum becomes an integral.

∑=

=n

iiiVqU

121

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V due to an infinite line charge

R

L line charge with charge density l

Gaussian cylinder

Remember Gauss’s Law

∫∫ ==⋅s

o

insiden

s

QdAEAdEε

∫∫∫ += EndcapsbarrelRs n dAEdAEdAE0

oRBarrelR

LRLEAEελπ == )2(

LQinside λ=

RkE

RE

R

oR

λεπ

λ

2

or2

=

=

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V due to an infinite line charge (continued)

�

dV = − E ⋅d l = −ER

ˆ R ⋅d l = −ERdR

RR

kV

R

RR

RRk

RdRk

dREVV

ref

ref

refref

ref

P

R

R

R

R RrefP

P

ref

P

ref

ln2

and 0.ln and 0ln

because or 0 chooset can'you :Problem

ln2

2

λ

λ

λ

=

∞<<∴+∞=∞−∞=

∞==

−=

−=

−=−

∫

∫

Where we define V = 0 at R = Rref

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electric field lines V1 = 100 V

V2 = 80 VV3 = 60 VV4 = 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