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Q. 67 - Excercises And Problems

Expert-verified
Found in: Page 690

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# 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 $\lambda =a|y|$, where a is a constant with the units $\mathrm{C}/{\mathrm{m}}^{2}$.a. Draw a graph of $\lambda versusy$ 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 $\frac{4Q}{{L}^{2}}\text{.}$

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

See the step by step solution

## 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 $\lambda =a|y|.$

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 $\lambda =a|y|.$

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,

$Q=2{\int }_{0}^{\frac{L}{2}}\lambda dy$

- $\lambda$ is the linear charge density.

- L is the length of the rod.

Substitute ay for $\lambda$in above equation.

$Q=2{\int }_{0}^{\frac{L}{2}}aydy$ \\

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

role="math" localid="1650099028854" $=a×\frac{{L}^{2}}{4}\phantom{\rule{0ex}{0ex}}a=\frac{4Q}{{L}^{2}}$

## 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 $\lambda =a|y|.$

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

- L is the length of the rod.

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

- $\lambda$ is the charge density.

Substitute $a|y|$ for $\lambda$ in above equation.

$E=\frac{1}{4\pi {\epsilon }_{0}}\frac{2a|y|}{x}\frac{L}{\sqrt{{L}^{2}+4{x}^{2}}}$