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EM-2.3-1 Review The electric field E(r) is a very special type of vector field ¾ For electrostatics, the CURL of E(r) = zero, i.e. The physical meaning of the curl of a vector field: For an arbitrary vector field A(r) , if × A(r)0 for all points in space, then the vector A(r) rotates , or shears in some manner in that region of space Curl of Whirlpool Field, × v (r) 0 Curl of shear Field × v (r) 0

2.3 Electric Potential

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Page 1: 2.3 Electric Potential

EM-2.3-1

ReviewThe electric field E(r) is a very special type of vector field

For electrostatics, the CURL of E(r) = zero, i.e.

The physical meaning of the curl of a vector field:For an arbitrary vector field A(r) , if ∇×

A(r)≠0 for all

points in space, then the vector A(r) rotates, or shears in some manner in that region of space

Curl of Whirlpool Field, ∇ ×

v (r) ≠ 0

Curl of shear Field∇ ×

v (r) ≠ 0

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Review

By use of Stokes’ Theorem

There are two implications (assuming E(r) ≠ 0 everywhere):1. everywhere (for arbitrary closed surface S).2. implies path independence of this (arbitrary)

closed contour, C.

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Electric potential

Define a scalar point function, V(r), known as the electric potential

(integral version)

Reference pointBy convention, the point r = Οref is taken to be a standard reference point of electric potential, V(r) where V (r = Οref ) = 0 (usually r = ∞).

SI Units of Electric Potential = Volts

If V (r)= Οref = 0 @ the reference point, then V(r) depends only on point r .

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Electric potential (conti.)

Electric potential difference between two points a & b

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Electric potential (conti.)

Thus

The fundamental theorem for gradients states that

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Electric potential (conti.)

The above equation is true for any end-points a & b (and any contour from a → b). Thus the two integrands mustbe equal

Now (for electrostatics):

Thus

So, for Electrostatic problems, ∇×

E(r) = 0 will always be true !

Knowing V(r) enables you to calculate E (r ) !!

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Why is E(r) specified as negative gradient of the electric potential?

Consider the point charge problem

In spherical-polar coordinates

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Why is E(r) specified as negative gradient of the electric potential? (conti.)

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Why is E(r) specified as negative gradient of the electric potential? (conti.)

Q=+e Q=-e

Radial outward Lines of E(r) Radial inward Lines of E(r)

V(r) for a point charge Q

By defining E(r) as the negative gradient, this simultaneously defines that lines of E point outward from (+) charge, and point inward for (-) charge.

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Why is E(r) specified as negative gradient of the electric potential? (conti.)

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Equipotentials: point charge

For a point charge, q, there exist “imaginary” surfaces –concentric spheres of varying radii r = R1 < R2 < R3 < …whose spherical surfaces are surfaces of constant potentialThese “imaginary” surfaces of constant potential are known as equipotential surfaces

E

E

E

E

E

EE

V1

V2

The equipotentials of constant V(r) are everywhere perpendicular to lines of E(r) !

+q

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Equipotentials: Arbitrary charge distribution

Charged metal

Consider a charged metal

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Electrostatic Potential and Superposition Principle

We have seen that, for any arbitrary electrostatic charge distributions:

Since

or

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Electrostatic Potential and Superposition Principle (conti.)

Integrate from a common reference point, a = Οref

Since

Therefore

Note that this is a scalar sum, not a vector sum!

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Example 2.7

A uniformly charged spherical (conducting) shell of radius, R, find the electric field.

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Example 2.7 (conti.)

Calculate V(r) from

use law of cosines

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Example 2.7 (conti.)

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Example 2.7 (conti.)

Note that

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Example 2.7 (conti.)

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Example 2.7 (conti.)

Then electric field

Thus

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POISSON’S EQUATION & LAPLACE’S EQUATION

Poisson’s equation

Laplace’s equation

If

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Cartesian Coordinates

Cylindrical Coordinates

Spherical Coordinates

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Typical electrostatic problem

Given charge distribution

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Typical electrostatic problem

Given V(r)

Given E(r)

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Typical electrostatic problem : Summary

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BOUNDARY CONDITIONSLet h → 0

Example 2.4

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E is discontinuous across a charged interface

Therefore

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Tangential components of E across a charged surface

Let h → 0

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Normal derivative of the potential V

Since

But Thus

Since

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V across a charged surface

Let h → 0