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A solid ball of radius R has a charge density $\rho$ given by $\rho$ = $\rho _{0}$ $(1-r/R)$ for 0 ≤ r ≤ R. The electric field outside the ball is :

Option 1)
$\frac{\rho _{0}R^{3}}{\epsilon _{0}r^{2}}$

Option 2)
$\frac{\rho _{0}R^{3}}{12\epsilon _{0}r^{2}}$

Option 3)
$\frac{4\rho _{0}R^{3}}{3\epsilon _{0}r^{2}}$

Option 4)
$\frac{3\rho _{0}R^{3}}{4\epsilon _{0}r^{2}}$

Answers (2)

best_answer

As we learnt

If P lies outside -

$E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}$ $V_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r}$

$E_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r^{2}}$ $V_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r}$

-

Total charge on sphere:

$q=\int_{0}^{R}\rho dv=\int_{0}^{R}\rho_{0}\left ( 1- \frac{r}{R}\right ).4\Pi r^{2}dr$

$q=\rho_{0}.4\Pi \left [ \int_{0}^{R}r^{2}dr-\int_{0}^{R}{\frac{r^{3}}{R}}dr \right ]$

$q=\rho_{0}.4\Pi\left ( \frac{R^{3}}{3}-\frac{R^{3}}{4} \right )=\rho_{0}\left ( 4\Pi \right )\frac{R^{3}}{12}$

$q=\frac{\rho_{0}.\Pi R^{3}}{3}$

$E=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}}=\frac{1}{4\pi \epsilon _{0}}.\frac{\rho _{0}\pi r^{3}}{3.r^{2}}$

$E=\frac{\rho _{0}R^{3}}{12\epsilon _{0}r^{2}}$

Option 1)

$\frac{\rho _{0}R^{3}}{\epsilon _{0}r^{2}}$

Option 2)

$\frac{\rho _{0}R^{3}}{12\epsilon _{0}r^{2}}$

Option 3)

$\frac{4\rho _{0}R^{3}}{3\epsilon _{0}r^{2}}$

Option 4)

$\frac{3\rho _{0}R^{3}}{4\epsilon _{0}r^{2}}$

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divya.saini

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