An electric field `E_(0) = 3xx10^(4) Vm^(-1)` is established between the plates `0.05m` apart, of a parallel plate capacitor. After removing the charging battery, an uncharged metal plate of thickness `t = 0.1 m` is inserted between capacitor plates. Find the `p.d.` across the capacitor, (i) before (ii) after the indroduction of plates (iii) what would be the `p.d` if a dielectric slab `(K = 2)` were inroduced of place of metal plate.

Answer

`1500 V, 1200 V, 1350 V`

Explanation

(i) `P.d` across the capacitance before metal plate was inserted
`V_(0) = E_(0) d = 3xx10^(4) xx0.05 = 1500 V`
(ii) After introduced of plate, `V = E_(0) (d - t)`
(ii) When dielectric slab is introduced, `K = 2`
`V = E_(0) (d - t) + (E_(0))/(K) t`
`= 1200 + (3xx10^(4)xx0.01)/(2) = 1350 V`