A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by `Q_0` , `V_0`, `E_0` and `U_0` respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one as

Options

A. `Q gt Q_(0)`
B. `V gt V_(0)`
C. `E gt E_(0)`
D. `U gt U_(0)`

Answer

A, D

Explanation

Since `Q_(0) = C_(0) V_(0)` and `Q = CV_(0) = epsilon_(r) C_(0) V_(0)`
So `(Q)/(Q_(0)) = in_(r), i.e., Q gt Q_(0)`
Now `U_(0) = (1)/(2) C_(0)V_(0)^(2), U = (1)/(2) CV^(2) = (1)/(2) epsilon_(r) C_(0) V^(2)`
`(U)/(U_(0)) = epsilon_(r), i.e., U gt U_(0)`
`V` and `E` remain same.