A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho = k\ r^{a}$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r = \frac{R}{2}$ is $\frac{1}{8}$ times that at $r = R,$ find the value of $a$

Question

Dr. Abubacker · Accepted Answer

(c) $\rho = K\ r^{a},E\left( r = \frac{R}{2} \right) = \frac{1}{8}E(r = R)$
$\frac{q_{enclosed}}{4\pi\varepsilon_{0}(R/2)^{2}} = \frac{1}{8}\frac{Q}{4\pi\varepsilon_{0}R^{2}}$
$32\ q_{enc.} = Q$
$q_{enc.} = \int_{0}^{R/2}{Kr^{a}\ 4\pi r^{2}\ dr = \frac{(4\pi K)}{a + 3}\left( \frac{R}{2} \right)^{a + 3}}$
$Q = \frac{4\pi K}{(a + 3)}R^{a + 3} \Rightarrow \frac{Q}{q_{enc.}} = 2^{a + 3} \Rightarrow 2^{a + 3} = 32$
$\Rightarrow a = 2$

Q. A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho = k\ r^{a}$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r = \frac{R}{2}$ is $\frac{1}{8}$ times that at $r = R,$ find the value of $a$

Explanation

Reviewed by Dr. Abubacker

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