Solveeit Logo

Question

Question: Two concentric conducting spherical shells having radii a and b are charged to q<sub>1</sub>& q<sub>...

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

A

q14πε0aq24πε0b\frac{q_{1}}{4\pi\varepsilon_{0}a} - \frac{q_{2}}{4\pi\varepsilon_{0}b}

B

q24πε0(1a1b)\frac{q_{2}}{4\pi\varepsilon_{0}}\left( \frac{1}{a} - \frac{1}{b} \right)

C

q14πε0(1a1b)\frac{q_{1}}{4\pi\varepsilon_{0}}\left( \frac{1}{a} - \frac{1}{b} \right)

D

None of these.

Answer

q14πε0(1a1b)\frac{q_{1}}{4\pi\varepsilon_{0}}\left( \frac{1}{a} - \frac{1}{b} \right)

Explanation

Solution

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

v1 =14πε0q1a+14πε0q2b\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}}{a} + \frac{1}{4\pi\varepsilon_{0}}\frac{q_{2}}{b} . . . . (1)

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

V2 = 14πε0q1b+14πε0q2b\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}}{b} + \frac{1}{4\pi\varepsilon_{0}}\frac{q_{2}}{b}

⇒ v = v1 – v2 ⇒ v = 14πε0q1a14πε0q1b\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}}{a} - \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}}{b}

⇒ v= q14πε0(1a1b)\frac{q_{1}}{4\pi\varepsilon_{0}}\left( \frac{1}{a} - \frac{1}{b} \right)