Question

If charge given to inner sphere is q then what should be charge on outer sphere so that potential energy of system is minimum?

Answer 1

Answer:

Potential at the surface of inner sphere =0 (grounded)

14πε0qr2+14πε0q'r1=0

⇒q'=−qr1r2

Answer 2

Potential energy :

Explanation:

The charge on the inner sphere is a q.

The charge on the outer sphere is a " q' ".

Let r1 is the radius of the inner circle.

Let r2 is the radius of the outer circle.

The potential energy of a sphere is calculated by the formula

$\displaystyle E = \frac{1}{4\pi \epsilon_0 }\frac{q}{r^2}$

The potential energy of an inner sphere is given by

$\displaystyle E_1 = \frac{1}{4\pi \epsilon_0 }\frac{q}{r_1^2}$

The potential energy of an inner sphere is given by

$\displaystyle E_2 = \frac{1}{4\pi \epsilon_0 }\frac{q'}{r_2^2}$

But the minimum potential energy of the sphere is 0.

i.e. E = 0

∴ E = E1 + E2

  0  =  $\displaystyle \frac{1}{4\pi \epsilon_0 }\frac{q}{r_1^2}$   $\displaystyle + \ \ \frac{1}{4\pi \epsilon_0 }\frac{q'}{r_2^2}$

⇒ $\displaystyle \mathbf{q' = \frac{-qr_2^2}{r_1^2}}$

∴ The charge on the outer sphere so that the potential energy of the system is minimum is $\displaystyle{q' = \frac{-qr_2^2}{r_1^2}}$