Two metalic balls of radii `R_(1)` and `R_(2)` are kept in vacuum at a large distance compared to their radii. Find the ratio of the charges on the two balls for which electrostatic energy of the system is minimum. What is the potential difference between the two balls for this ratio ? Total charge of the balls is constant. Neglect the interaction energy. (Charge distribution on each ball is uniform)

Answer

`Q_(1)/Q_(2)=R_(1)/R_(2)`

Explanation

Electrostatic energy of sytem = Interaction Energy+ self Energy
(Let the total charge of balls be Q)
`U=0+(KQ_(1)^(2))/(2R_(1))+(K(Q-Q_(1))^(2))/(2R_(2))`
here for it's minimum value `(dU)/(dQ_(1))=0`
`=K[(2Q_(1))/(2R_(1))+(2(Q-Q_(1)))/(2R_(2))(0-1)]=0`
`rArr Q_(1)/R_(1)-Q_(2)/R_(2)=0 [ :' Q-Q_(1)=Q_(2)]rArr Q_(1)/Q_(2)=R_(1)/R_(2)`