A hollow charged metal sphere has radius $r$. If the potential difference between its surface and a point at a distance $3r$ from the centre is$\ V$, then electric field intensity at a distance $3r$ is

Question

Dr. Abubacker · Accepted Answer

(c) Here $V = \frac{q}{4\pi\varepsilon_{0}.r} - \frac{q}{\begin{array}{r}
4\pi\varepsilon_{0}(3r) \
\ 
\end{array}} = \frac{1}{4\pi\varepsilon_{0}}.\frac{2q}{3r}\ \ $and
$E = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{(3r)^{2}} = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{9r^{2}}$
On simplification, we get $\ \frac{E}{V} = \frac{1}{6r}\ or\ E = \frac{V}{6r}$

Q. A hollow charged metal sphere has radius $r$. If the potential difference between its surface and a point at a distance $3r$ from the centre is$\ V$, then electric field intensity at a distance $3r$ is

Explanation

Reviewed by Dr. Abubacker

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