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) Energy

B) Voltage

C) Electric Field

D) Charge

Asked by lissyjeslyn | 11th Feb, 2020, 11:39: AM

Expert Answer:

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×V2
after inserting dielectric, energy available in the capacitor = (1/2)×(kC)×(V/k)= 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 
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. 

Answered by Shiwani Sawant | 11th Feb, 2020, 01:10: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.