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A parallel plate air capacitor is connec...

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

A

(a) `QgtQ_0`

B

(b) `VgtV_0`

C

(c) `EgtE_0`

D

(d) `UgtU_0`

Text Solution

Verified by Experts

The correct Answer is:
A, D

(i) `P.d. = V_0`
Capacitance =C
(ii) `Q_0=CV_0`
(iii) Potential Energy
`=1/2CV_0^2`
(iv) `E=(V_0)/(d)`
`P.d.=V_0`
Capacitance =KC
[K is the dielectric constant of slab Kgt1]
New charge `=KCV_0`
New potential energy
`=1/2KCV_0^2`
`E=(V_0)/(d)`
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