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  • A spherically-symmetric charge distribution is considered with charge density varying as <img alt="\rho(r)= \begin{cases}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for }

A spherically-symmetric charge distribution is considered with charge density varying as
\rho(r)= \begin{cases}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { zero } & \text { for } r>R\end{cases}
Where, \mathrm{r(r<R)} is the distance from the centre \mathrm{O} (as shown in the figure). The electric field at point \mathrm{P} will be :
 

Option: 1

\frac{\rho_{0} \mathrm{r}}{4 \varepsilon_{0}}\left(\frac{3}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)


Option: 2

\frac{\rho_{0} \mathrm{r}}{3 \varepsilon_{0}}\left(\frac{3}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)


Option: 3

\frac{\rho_{0} \mathrm{r}}{4 \varepsilon_{0}}\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)


Option: 4

\frac{\rho_{0} \mathrm{r}}{5 \varepsilon_{0}}\left(1-\frac{\mathrm{r}}{\mathrm{R}}\right)


Answers (1)

best_answer

\mathrm{\rho \left ( r \right )=\rho _{o}\left ( \frac{3}{4}-\frac{r}{R} \right )For \: \: \: r\leq R}

\mathrm{=Zero}                           \mathrm{r> R}

Change in the integral element is,

\mathrm{dq=\rho 4\pi x^{2}dx}

\mathrm{d q =\rho_0\left(\frac{3}{4}-\frac{x}{R}\right) \times 4 \pi x^2 d x }

\mathrm{q=\int d q =\rho _0 \times 4 \pi \int_0^r\left(\frac{3}{4} x^2 d x-\frac{x^3}{R} d x\right) }

   \mathrm{=4 \pi \rho _0\left(\frac{3}{4}\left[\frac{x^3}{3}\right]_0^r-\left[\frac{x^4}{4 R}\right]_0^r\right) }

\mathrm{q=4 \pi \rho_0 r^3\left[\frac{1}{4}-\frac{r}{4 R}\right) }

By  Gauss's law,

\mathrm{\oint \bar{E}.d\bar{A}=\frac{q_{enc}}{\varepsilon _{0}}}

\mathrm{E \times \left ( 4\pi r^{2} \right )=\frac{4\pi \rho _{o}r^{3}}{4}\left ( 1-\frac{r}{R} \right )}

\mathrm{E=\frac{\rho_0 r}{4\varepsilon _{o}}\left ( 1-\frac{r}{R} \right )}

Hence 3 is the correct option.
 

Posted by

Pankaj Sanodiya

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