A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Options

A. zero
B. `(1)/(2) (K - 1) CV^(2)`
C. `(CV^(2) (K - 1))/(K)`
D. `(K - 1) CV^(2)`

Answer

A

Explanation

Initial energy `U_(i) = (Q^(2))/(2 K C)`
When the slab is removed, `U = (Q^(2))/(2C)`,
When the slab is removed, `U = (Q^(2))/(2C)`,
When dielectric slab is introduced, energy
becomes `U_(f) = (Q^(2))/(2 KC) = U_(i)`.
`:.` Net work done by the system `= U_(f) - U_(i)`
= Zero