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?
C) Electric Field
Asked by lissyjeslyn | 11th Feb, 2020, 11:39: AM
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)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
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
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