The plates of a capacitor are connected to a source of `e.m.f 320 V`. Between the platesm there are two dielectric slabs each of uniform thickness filling the whole space. One slab `A` is of thickness `4 mm` and dielectric constant `8`. The other slab `B` is of thickness `3 mm` and dielectric constant `10`.

Options

A. The electric field intensity in `A` is `5 xx 10^(4) N//C`
B. The electric field intensity in `B` is `4 xx 10^(4) N//C`
C. The energy stored per unit area of the capacitor is `5.7 xx 10^(-4) J`.
D. The (induced) bound charge on `B` is more than in `A`.

Answer

A, B, C, D

Explanation

`(E)/(8)xx4xx10^(-3)+(E)/(10)xx 2xx 10^(-3) =320 or E=4 xx 10^(5)`
Therefore `(E)/(8)= 5 xx 10^(4) N//C (E)/(10)=4 xx 10^(4) N//C`
Energy stored per unit area `=(1)/(2)cv^(2)=5.7 xx10^(-4) J sigma(on B)` is greater than `sigma (on A) because K_(B) gt K_(A)`.\