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 the electric field intensity at distance 3r 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 =$\frac { \mathrm { kq } } { \mathrm { r } }$– $\frac { \mathrm { kq } } { 3 \mathrm { r } }$ = V

$\frac { \mathrm { kq } } { \mathrm { r } }$ × $\frac { 2 } { 3 }$= V

E_B =$\frac { \mathrm { kq } } { ( 3 \mathrm { r } ) ^ { 2 } }$= $\frac { 3 } { 2 }$ =$\frac { 3 \mathrm { Vr } } { 2 \times 9 \times \mathrm { r } ^ { 2 } }$=