Question

A non-conducting sphere of radius $R=5\text{cm}$ has its centre at the origin $O$ as shown and have two spherical cavities of radius $r=1\text{cm}$ whose centres are at $(0,3\text{cm})$ and $(0,-3\text{cm})$. If solid material of the sphere has uniform positive charge density $\rho =\frac{1}{\pi }\mu {\text{C-m}}^{-3}$, then potential at point $\text{P}(4\text{cm},0)$ is

• A. $35\text{V}$
• B. $70\text{V}$
• C. $-35\text{V}$
• D. $-70\text{V}$

Answer 1

The correct option is A $35\text{V}$

Given:
$R=5\text{cm}=5\times {10}^{-2}\text{m};$
$r=1\text{cm}={10}^{-2}\text{m};$
$\rho =\frac{1}{\pi }\mu {\text{C-m}}^{-3}=\frac{1}{\pi }\times {10}^{-6}{\text{C-m}}^{-3}$

Charge on the sphere is

$Q=\frac{4\pi }{3}{R}^3\rho$

$\Rightarrow Q=\frac{4}{3}\pi \times {(5\times {10}^{-2})}^3\times \frac{1}{\pi }\times {10}^{-6}$

$\Rightarrow Q=\frac{500}{3}\times {10}^{-12}\text{C}$

Charge accumulated in the volume of cavity is

$q=\frac{4}{3}\pi r^3\rho$

$\Rightarrow q=\frac{4}{3}\pi \times {(1\times {10}^{-2})}^3\times \frac{1}{\pi }\times {10}^{-6}$

$\Rightarrow q=\frac{4}{3}\times {10}^{-12}\text{C}$

Now potential at point $\text{P}$

$V_P=V_{\text{solid sphere}}-2V_{\text{cavity}}$

$\Rightarrow V_P=\frac{kQ}{2R}\{3-\frac{x^2}{R^2}\}-2\frac{Kq}{r}$

Here, $x=4\text{cm}=4\times {10}^{-2}\text{m}.$

$V_p=\frac{9\times {10}^9\times \left(500/3\right)\times {10}^{-12}}{2\times 5\times {10}^{-2}}\{3-{\left(\frac{4}{5}\right)}^2\}-\frac{2\times 9\times {10}^9\times \left(4/3\right)\times {10}^{-12}}{5\times {10}^{-2}}$

$\Rightarrow V_p=\frac{873}{25}=34.92\approx 35\text{V}$

So, option $\left(a\right)$ is correct.

Why this question? Concept: Electric potential inside a uniformly charged sphere is $V=\frac{kQ}{2R}\{3-\frac{r^2}{R^2}\}(\text{For}r<R)$