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W04D1 Electric Potential and Gauss’ Law

Equipotential Lines

Today’s Reading Assignment

Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.5

Announcements

Exam One Thursday Feb 28 7:30-9:30 pm Room Assignments (See Stellar webpage announcements)

Review Tuesday Feb 26 from 9-11 pm in 26-152

PS 3 due Tuesday Tues Feb 26 at 9 pm in boxes outside 32-082 or 26-152

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Outline

Continuous Charge Distributions

Review E and V

Deriving E from V

Using Gauss’s Law to find V from E

Equipotential Surfaces

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Continuous Charge Distributions

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Continuous Charge DistributionsBreak distribution into

infinitesimal charged elements of charge dq. Electric Potential difference between infinity and P due to dq.

Superposition Principle:

Reference Point:

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Group Problem

Consider a uniformly charged ring with total charge Q. Find the electric potential difference between infinity and a point P along the symmetric axis a distance z from the center of the ring.

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Group Problem: Charged RingChoose

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Electric Potential and Electric Field

Set of discrete charges:

Continuous charges:

If you already know electric field (Gauss’ Law)

compute electric potential difference using

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Using Gauss’s Law to findElectric Potential from Electric Field

If the charge distribution has a lot of symmetry, we use Gauss’s Law to calculate the electric field and then calculate the electric potential V using

B

B A

A

V V d E s

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Group Problem: Coaxial Cylinders

A very long thin uniformly charged cylindrical shell (length h and radius a) carrying a positive charge +Q is surrounded by a thin uniformly charged cylindrical shell (length h and radius a ) with negative charge -Q , as shown in the figure. You may ignore edge effects. Find V(b) – V(a).

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Worked Example: Spherical Shells

These two spherical shells have equal but opposite charge.

Find

for the regions

(i) b < r

(ii) a < r < b

(iii) 0 < r < a

Choose

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Electric Potential for Nested Shells

From Gauss’s Law

Use

Region 1: r > b

r

No field No change in V!

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Electric Potential for Nested Shells

Region 2: a < r < b

r

Electric field is just a point charge.

Electric potential is DIFFERENT – surroundings matter

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Electric Potential for Nested Shells

Region 3: r < a

r

Again, potential is CONSTANT since E = 0, but the potential is NOT ZERO for r < a.

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Group Problem: Charge Slab

Infinite slab of thickness 2d, centered at x = 0 with uniform charge density . Find

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Deriving E from V

A = (x,y,z), B=(x+Δx,y,z)

Ex = Rate of change in V with y and z held constant

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Gradient (del) operator:

If we do all coordinates:

Deriving E from V

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Concept Question: E from VConsider the point-like charged objects arranged in the figure below. The electric potential difference between the point P and infinity and is

From that can you derive E(P)?

1. Yes, its kQ/a2 (up)

2. Yes, its kQ/a2 (down)

3. Yes in theory, but I don’t know how to take a gradient

4. No, you can’t get E(P) from V(P)

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Concept Question Answer: E from V

The electric field is the gradient (spatial derivative) of the potential. Knowing the potential at a single point tells you nothing about its derivative.

People commonly make the mistake of trying to do this. Don’t!

4. No, you can’t get E(P) from V(P)

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Group Problem: E from VConsider two point like charged objects with charge –Q located at the origin and +Q located at the point (0,a).

(a)Find the electric potential V(x,y)at the point P located at (x,y).

(b)Find the x-and y-components of the electric field at the point P using

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Concept Question: E from V

The graph above shows a potential V as a function of x. The magnitude of the electric field for x > 0 is

1. larger than that for x < 0 2. smaller than that for x < 0 3. equal to that for x < 0

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Concept Question Answer: E from V

The slope is smaller for x > 0 than x < 0Translation: The hill is steeper on the left than on the right.

Answer: 2. The magnitude of the electric field for x > 0 is smaller than that for x < 0

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Concept Question: E from V

The above shows potential V(x). Which is true?

1. Ex > 0 is positive and Ex < 0 is positive

2. Ex > 0 is positive and Ex < 0 is negative

3. Ex > 0 is negative and Ex < 0 is negative

4. Ex > 0 is negative and Ex < 0 is positive

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Concept Question Answer: E from V

E is the negative slope of the potential, positive on the right, negative on the left,

Translation: “Downhill” is to the left on the left and to the right on the right.

Answer: 2. Ex > 0 is positive and Ex < 0 is negative

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Group Problem: E from V

A potential V(x,y,z) is plotted above. It does not depend on x or y.

What is the electric field everywhere?

Are there charges anywhere? What sign?

-5 0 50

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Pot

entia

l (V

)

Z Position (mm)

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Equipotentials

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Topographic Maps

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Equipotential Curves: Two Dimensions

All points on equipotential curve are at same potential.Each curve represented by V(x,y) = constant

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Direction of Electric Field E

E is perpendicular to all equipotentials

Constant E field Point Charge Electric dipole

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Direction of Electric Field E

http://web.mit.edu/viz/EM/visualizations/electrostatics/InteractingCharges/zoo/zoo.htm

E is perpendicular to all equipotentials

Field of 4 charges Equipotentials of 4 charges

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Properties of Equipotentials

E field lines point from high to low potential

E field lines perpendicular to equipotentials

E field has no component along equipotential

The electrostatic force does zero work to move a charged particle along equipotential