Question

Two conducting spheres having radii a and b are charged to q₁&q₂ respectively. The potential difference between 1 & 2 will be

• A. $\frac { \mathrm { q } _ { 1 } } { 4 \pi \varepsilon _ { 0 } \mathrm { a } }$ $- \frac { \mathrm { q } _ { 2 } } { 4 \pi \varepsilon _ { 0 } \mathrm {~b} }$
• B. $\frac { \mathrm { q } _ { 2 } } { 4 \pi \varepsilon _ { 0 } }$ $\left( \frac { 1 } { a } - \frac { 1 } { b } \right)$
• C.
• D. None of these

Answer 1

Answer: (C)

Answer 2

The potential on the surface of the sphere 1 is given by,

V_{1 =} $\frac { 1 } { 4 \pi \varepsilon _ { 0 } } \frac { \mathrm { q } _ { 1 } } { \mathrm { a } } + \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \frac { \mathrm { q } _ { 2 } } { \mathrm {~b} }$

The potential on the surface of the sphere 2 is given by

V₂$\frac { 1 } { 4 \pi \varepsilon _ { 0 } } \frac { \mathrm { q } _ { 1 } } { \mathrm {~b} } + \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \frac { \mathrm { q } _ { 2 } } { \mathrm {~b} }$

∴ Potential difference V = V₁ - V₂ =

⇒ V =