Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

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Short Answer

A rod of length L lies along the y-axis with its center at the ALC origin. The rod has a nonuniform linear charge density $λ = a | y |$ , where a is a constant with the units $C / m^{2}$ .
a. Draw a graph of $λ v e r s u s y$ over the length of the rod.
b. Determine the constant a in terms of L and the rod's total charge Q.
c. Find the electric field strength of the rod at distance x on the x-axis.

a. the graph \lambda \text { versus } y \text { over the length of the rod is shown in figure

b. the constant a in terms of L is $\frac{4 Q}{L^{2}} .$

c. the electric field strength of the rod at a distance x on the x -axis is $\frac{1}{4 π ε_{0}} \frac{2 a | y |}{x} \frac{L}{\sqrt{L^{2} + 4 x^{2}}} .$

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part (a) step 1 : given information

A rod of length L lies along the y-axis with its center at the origin. The rod has a non uniform linear charge density $λ = a | y | .$

Figure I

part (b) step 1 : given information

Given info: A rod of length L lies along the y-axis with its center at the origin. The rod has a non uniform linear charge density $λ = a | y | .$

The total charge is twice the integral of the linear charge density from 0 to $\frac{L}{2}$

So, total charge on the rod is,

$Q = 2 \int_{0}^{\frac{L}{2}} λ d y$

- $λ$ is the linear charge density.

- L is the length of the rod.

Substitute ay for $λ$ in above equation.

$Q = 2 \int_{0}^{\frac{L}{2}} a y d y$ \\

$= 2 a {|\frac{y^{2}}{2}|}_{0}^{\frac{L}{2}}$

role="math" localid="1650099028854"

= a \times \frac{L^{2}}{4} a = \frac{4 Q}{L^{2}}

part (c) step 1 : given information

Given info: A rod of length L lies along the y-axis with its center at the origin. The rod has a non uniform linear charge density $λ = a | y | .$

The electric field strength of a finite rod at distance x and non uniform charge density is, $E = \frac{1}{4 π ε_{0}} \frac{2 λ}{x} \frac{1}{\sqrt{1 + \frac{4 x^{2}}{L^{2}}}}$

- L is the length of the rod.

- x is the distance covered by rod on x-axis.

- $λ$ is the charge density.

Substitute $a | y |$ for $λ$ in above equation.

E = \frac{1}{4 π ε_{0}} \frac{2 a | y |}{x} \frac{L}{\sqrt{L^{2} + 4 x^{2}}}

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