Question

If radius of a hollow metallic sphere is $R$. If the potential difference between it's surface and a point at a distance $3R$ from it's centre is $V$ then the electric field intensity at a distance $3R$ from it's centre will be

• A. $\frac{V}{2R}$
• B. $\frac{V}{3R}$
• C. $\frac{V}{4R}$
• D. $\frac{V}{6R}$

Answer 1

The correct option is D $\frac{V}{6R}$


Referring to the figure , the potential difference $\left(V\right)$ between the surface and at a point $P$
$V=V_s-V_p$

Substituting the value of $V_s$ and $V_P$, we get
$V=\frac{q}{4\pi {ϵ}_0}[\frac{1}{R}-\frac{1}{3R}]$

$\Rightarrow V=\frac{q}{4\pi {ϵ}_0}\left[\frac{2}{3R}\right]........\left(1\right)$

Let, $E=$ Electric field at $\text{P}$,

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{(3R{)}^2}$

$\Rightarrow E=\left(\frac{1}{4\pi {ϵ}_0}\frac{q}{3R}\right)\frac{1}{3R}.....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get

$E=\frac{V}{2\times 3R}$

$\therefore E=\frac{V}{6R}$

Hence, option (d) is the correct answer.