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Q. 67 - Excercises And Problems

Expert-verifiedFound in: Page 690

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 $\lambda =a\left|y\right|$, 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 $\frac{1}{4\pi {\epsilon}_{0}}\frac{2a\left|y\right|}{x}\frac{L}{\sqrt{{L}^{2}+4{x}^{2}}}\text{.}$

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 $\lambda =a\left|y\right|.$

Figure I

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 $\lambda =a\left|y\right|.$

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" $=a\times \frac{{L}^{2}}{4}\phantom{\rule{0ex}{0ex}}a=\frac{4Q}{{L}^{2}}$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 $\lambda =a\left|y\right|.$

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 $a\left|y\right|$ for $\lambda $ in above equation.

$E=\frac{1}{4\pi {\epsilon}_{0}}\frac{2a\left|y\right|}{x}\frac{L}{\sqrt{{L}^{2}+4{x}^{2}}}$94% of StudySmarter users get better grades.

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