Question 324: Electrostatics & Capacitance

Question

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

Accepted Answer

$$\left( {4\pi {r^2}} \right){E_\rho } = {{{Q_{in}}} \over {{\varepsilon _0}}}$$
$$ = {{\int_0^r {{\rho _0}\left( {{3 \over 4} - {r \over 4}} \right)4\pi {r^2}dr} } \over {{\varepsilon _0}}}$$
$$ = {{{\rho _0}\pi 4} \over {{\varepsilon _0}}}\left( {{{{r^3}} \over 4} - {{{r^4}} \over {4R}}} \right)$$
$${E_\rho } = {{{\rho _0}} \over {4{\varepsilon _0}}}\left( {r - {{{r^2}} \over R}} \right)$$
$$ = {{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {1 - {r \over R}} \right)$$

Answer

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

Answer

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

Answer

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

Answer

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

Electrostatics & Capacitance

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