Question

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 distance 3 r from the centre is -

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

Answer 1

Answer: (C)

$\frac{V}{6r}$

Answer 2

V_A – V_B = $\frac{KQ}{r}$ – $\frac{KQ}{3r}$

= $\frac{KQ2}{3r}$, V = $\frac{2}{3}\frac{KQ}{r}$

E = $\frac{KQ}{(3r)^{2}}$ = $\frac{KQ}{9r^{2}}$ = $\frac{3}{2}\frac{rV}{9r^{2}}$\ E = $\frac{V}{6r}$