Question: Q. 26. (i) Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ’ C ’ and having charge ’ Q ‘.

(ii) How will the (a) energy stored and (b) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ’ K ’ ?

U] [O.D. I, II, III 2012]

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Solution:

Ans. (i) Expression of electrostatic energy stored in a capacitor

(ii) Effect on energy stored and electric field in capacitor due to dielectric material. Potential difference between the plates of capacitor

V=q/C

Work done in adding an additional charge dq on the capacitor

dW=V×dq =(qC)×dq

Total energy stored in the capacitor,

(1)U=d W=0QqCdq=12Q2C

When battery is disconnected:

(a) Energy stored will be decreased or energy stored =1κ times the initial energy.

1/2

(b) Electric field would decrease.

[CBSE Marking Scheme 2012]

Long Answer Type Questions



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