# An air-core capacitor is charged by a battery. After disconnecting it from the battery, a dielectric slab is fully inserted in between its plates. Now, which of the following quantities remains constant?A) EnergyB) VoltageC) Electric FieldD) Charge

### After disconnecting battery let us assume the charge on the plates is Q and the potential difference is V. The relation connecting capacitance C, charge Q and potential difference V is given by C = Q/V.when dielectric material is inserted between parallel plates the capacitance increases and the new capacitance C' is given by C' = k×C, where k is dielectric constant. Charge Q remains same. Hence potential difference reduces and the new value of potential difference is V/k .Energy stored in the capacitor before inserting dielectric  = (1/2)×C×V2after inserting dielectric, energy available in the capacitor = (1/2)×(kC)×(V/k)2 = k×(1/2)×C×V2

Electric field is given as
E = V/d
But as there is a change in the potential difference the new value of electric field will become
E = V/kd
Thus,
from analysing the data given above we can say that only charge will remain constant. All other quantities - energy, electric field, capacitance and potential difference will vary.

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