Question

A parallel plate capacitor has a capacitance of `2 mu F`. A slab of dielectric constant 5 is inserted between the plates and the capacitor is charged to `100 V` and then isolated . (a) What is the new potential diff., if the dielectric slab is removed ? (b) How much work is required to remove teh dielectric slab ?

Answer 1

Correct Answer - `500 V; 0.2 J`
Here, `C_(0) = 2 muF = 2xx10^(-6)F, K = 5, V = 100` volt
When dielectric slab is removed, capacity becomes `(1)/(K)` times and potential becomes `K` times.
`:.` New potential `V_(0) = 5xx100 = 500 V`
Work required to remove dielectric
`= U_(2) - U_(1) = (1)/(2) C_(0) V_(0)^(2) - (1)/(2) CV^(2)`
`= (1)/(2) [2xx10^(-6) (500)^(2) - 10xx10^(-6) xx (100)^(2)]`
`= (1)/(2) (0.5- 0.1) = 0.2 J`