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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

`QgtQ_0`

B

`VgtV_0`

C

`EgtE_0`

D

`UgtU_0`

Text Solution

Verified by Experts


Before introducing dielectric slab:
Potential difference `=V_(0)`
Capacitance `=C`
charge `q_(0)=CV_(0)`
`PE,U_(0)=(1)/(2)CV_(0)^(2)`
After introduction of dielectric slab:
Potential difference `=V_(0)`
Capacitance `=KC[K` is the dielectric constant of slab `K gt1`] ltbr New charge `,Q=KCV_(0)`
New `PE,U=(1)/(2)KCV_(0)^(2)`
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