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Introduction to Electrodynamics
Found in: Page 83
Introduction to Electrodynamics

Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

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Short Answer

For the configuration of Prob. 2.16, find the potential difference between a point on the axis and a point on the outer cylinder. Note that it is not necessary to commit yourself to a particular reference point, if you use Eq. 2.22.

The potential difference ispa24ε01-2Inab.

See the step by step solution

Step by Step Solution

Step 1: Define electric potential.

The line integral of electric field is known electric potential. The electric potential is described as,

V(r)=-0rE×dl

Here, V(r) is the potential at position r , E is the electric field, and is the length of small element.

Consider the two points are a and b.

The electric potential difference between two points determined as,

V(b)-V(b)=-0bE-dl+0aE.dl =-0bE-dl-a0E.dl =-abE-dl

Here, V(a) is the potential at position a.

V(b) is the potential at position b.

Step 2: Determine electric field configuration

Write the configuration of electric field for cylinder.

Es=pS2ε0s<apa22sε0a<s<b0,s>b

Here, a is the inner radius of the cylinder, b is the outer radius of the radius of sphere, s is the radius of the Gaussian surface, p is the volume charge density and ε0 is the permittivity of the free space.

Step 3: Determine electric potential.

The electric potential is,

V=Vb-Va =-abE(s)ds

Now, substituting the value of electric field in above equation,

V(b)-V(a)=-s-bs-apa22sε0ds-s-as-0ps2ε0ds-0

Solve for the voltage difference as,

V(b)-V(a)=-s-bs-apa22sε0ds--s-as-0ps2ε0ds = -pa22ε0s-bs-a1sds-p2ε0s-as-0ds =-pa22ε0InS|ba-p2ε0s22a0

Further solving above equation,

V(b)-V(a)=-pa22ε0(In a -In b)-p2ε00-a22 =22pa22ε0In ab+p2ε0a22 =2pa24ε0In ab+p4ε0(a2) = pa24ε0 1-2 In abThus, the potential difference is pa24ε0 1-2 In ab.

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