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

Answer

C

Explanation

When the dielectric slab is introduced, C is increased. `therefore C gtC_(0)`
Bu as the battery is still connected, V remains constant,
and `because Q+CV,Q gt Q_(0)` as `C gtC_(0)`
and the energy `U=(1)/(2)CV^(2)` as `C gt C_(0),U gt U_(0)`
Similarly `E=(V)/(d)` but V and d are not changed and
`U lt U_(0)`
`therefore E=E_(0)`. thus `E gt E_(0)` and `V gt V_(0)` are wrong.
`Q gt Q_(0)` is the correct option