Answer :

Option b) is correct because when a dielectric slab W is inserted in the capacitor in the presence of a battery the capacitance increases by a factor of Kdielectric constant)

Where

K is the dielectric constant

ε_{o} is the permittivity of the free space

A is the area of the plate,

d is the distance between the plates of the capacitor,

**As the capacitance increases with the insertion of the dielectric, the charge appearing on the capacitor increases**

Since,

Q=CV

Where,

Q is the charge on the capacitor

C is the capacitance of the capacitor

V is the potential difference supplied by the battery

Whereas capacitance does not change in case of inserting slab after removing the battery.

Optionc) is correct as

In process WXY after inserting a dielectric slab in the capacitor, the capacitance becomes

Where C_{0} is the capacitance in a vacuum and K is the dielectric constant. Now if the capacitor is connected to the battery of emf ϵ, then potential difference across the capacitor is given by ϵ, and the stored electrical energy is given by

Whereas in process XYW the energy is given by

**Therefore, The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.**

Option→d) is correct because in both cases Electric field in the capacitor reduces to

Where

E_{0}=electric field in c=vacuum

K=dielectric constant

note that it does not matter whether the battery is connected afterwards or before in 4^{th} part)

**The electric field in the capacitor after the action XW is the same as that after WX.**

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