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The volume charge density in a spherical ball of radius $R$ varies with distance $r$ from the centre as $\rho(r)=\rho_0\left[1-\left(\frac{r}{R}\right)^3\right]$, where, $\rho_0$ is a constant. The radius at which the field would be maximum is
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Verified Answer
The correct answer is:
$\frac{R}{2^{1 / 3}}$
Given, $\quad \rho(r)=\rho_0\left[1-\left(\frac{r}{R}\right)^3\right]$
So, effective charge at $r$ is $q(r)=\int \rho(r) \cdot V$
$$
=\rho_0\left[1-\left(\frac{r}{R}\right)^3\right] \frac{4}{3} \pi r^3
$$
For maximum value of field,
$$
\begin{array}{rlrl}
& \frac{d q(r)}{d r} & =0 \\
\Rightarrow & \frac{4}{3} \pi \rho_0\left[3 r^2\left(1-\frac{r^3}{R^3}\right)+r^3\left(-\frac{3 r^2}{R^3}\right)\right] & =0 \\
\Rightarrow & & 3 r^2-\frac{3 r^5}{R^3}-\frac{3 r^5}{R^3} & =0 \\
\Rightarrow & r^3 & =\frac{R^3}{2} \Rightarrow r & =\frac{R}{2^{1 / 3}}
\end{array}
$$
So, effective charge at $r$ is $q(r)=\int \rho(r) \cdot V$
$$
=\rho_0\left[1-\left(\frac{r}{R}\right)^3\right] \frac{4}{3} \pi r^3
$$
For maximum value of field,
$$
\begin{array}{rlrl}
& \frac{d q(r)}{d r} & =0 \\
\Rightarrow & \frac{4}{3} \pi \rho_0\left[3 r^2\left(1-\frac{r^3}{R^3}\right)+r^3\left(-\frac{3 r^2}{R^3}\right)\right] & =0 \\
\Rightarrow & & 3 r^2-\frac{3 r^5}{R^3}-\frac{3 r^5}{R^3} & =0 \\
\Rightarrow & r^3 & =\frac{R^3}{2} \Rightarrow r & =\frac{R}{2^{1 / 3}}
\end{array}
$$
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