A sperical shell of radius `R_(1)` with a unifrom charge `q` has a point charge `q_(0)` at the its centre . Find the work perfomed by the electric forces during the shell expansion form radius `R_(1)` to radius `R_(2)`.

Explanation

Iniitally, energy of the system, `U_(i) = W_(i) + W_(12)` where, `W_(1)` is the self energy and `W_(12)` is the normal energy.
So, `U_(i) = (1)/(2) (q^(2))/(4pi epsilon_(0) R_(1)) + (q q_(0))/(4pi epsilon_(0) R_(1))`
and on expansion, energy of the system,
`U_(f) = W'_(1) + W'_(12)`
`= (1)/(2) (q^(2))/(4pi epsilon_(0) R_(2)) + (q q_(0))/(4pi epsilon_(0) R_(2))`
Now, work doen by the field force, A equals the decrement in the electrical energy,
`i.e., A = (U_(i) - U_(f)) = (q (q_(0) + q//2))/(4pi epsilon_(0)) ((1)/(R_(1)) - (1)/(R_(2)))`
Alternate : The work of electric forces is equal to the decrease in electric energy of the system,
`A = U_(i) - U_(f)`
In order to find the difference `U_(i) - U_(f)` we note that upon expression of the shell, the electirc field and hence the energy localized in it, changed only in the hatched spherical layer consequently (Fig).
`U_(i) - U_(f) = int_(R_(1))^(R_(2)) (epsilon_(0))/(2) (E_(1)^(2) - E_(2)^(2)) . 4pi r^(2) dr`.
where `E_(1)` and `E_(2)` are the field intensties (in the bathched region at a distance `r` from teh centre of the system) before and after the expansion of the shell. By using Gauss's theorem, we find
`E_(i) = (1)/(4pi epsilon_(0)) (q + q_(0))/(r^(2))` and `E_(2) = (1)/(4pi epsilon_(0)) (q_(0))/(r^(2))`
As a result of intergation, we obtains
`A = (q (q_(0) + q//2))/(r^(2)) ((1)/(R_(1)) - (1)/(R_(2)))`