The capacity of a condenser increases both, when a conducting slab or an insulating slab is introduced between the plates of the condenser. In the former case, electric field `E = 0` inside the conductor and in the latter case, `E lt E_(0)`, inside the insulator. Thus, potentai difference `V = E xx d` decreases and hence capacity `C = Q//V` increases.
It should be clearly understood that when a dielectric slab is introduced inbetween the plates of a charged capacitor with battery connected across the plates,
(i) Capacity `C` increases, (ii) Potential `V` remains constant, (iii) Charge `Q = CV`, increases, (iv) Electric field `E` decreases,
(v) Energy stored `U = (1)/(2) CV^(2)` increases.
However, when battery across the plates of charged capacitor is put off and dielectric slab is introduced inbetween th plates of the capacitor, (i) Capacity `C` increases,
(ii) charge `Q` remains constant,
(iii) Potential `V = (Q)/(C )` decreases, (iv) Electric field. `E = V xx d` decreases, (v) Energy stored `U = (Q^(2))/(2C)` decreases.
Consider a parallel plate air capacitor with area of each plate `= 150 cm^(2)` and distance between its plates `= 0.8mm`. With the help of the passage given above, choose the most appropriate for each of the following questions :
The air capacitor is charged to `1200 V` and then filled with dielectric of `K = 3`. The charge on the plates will be

Options

A. `1.66xx10^(2) C`
B. `1.66xx10^(-10) C`
C. `1.99xx10^(7)C`
D. `1.99xx10^(-7)C`

Answer

D

Explanation

`Q = C_(0) V_(0) = 1.66xx10^(-10)xx1200`
`= 1.99xx10^(7) C`
On filling with dielectric after charging, the charge does not change. It remains the same.