A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the 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. `QgtQ_(0)`
B. `VgtV_(0)`
C. `EgtE_(0)`
D. `UgtU_(0)`

Answer

A, D

Explanation

When electric stab is introduced capacity gets increased while potential difference remains unchanged
`V=V_0,CgtC_0 , Q=CV therefore QgtQ_0`
`U=1/2CV^(2) therefore ugtU_0, E=v/d` but v and d both are unchanged .
Therefore ,`E=E_0` , Therefore correct option are (A) and (D).