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Capacitor

JEE Physics · 62 questions · Page 7 of 7 · Click an option or "Show Solution" to reveal answer

Q61
A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be
A 1/21/2
B 11
C 22
D 1/41/4
Correct Answer
Option A
Solution

Required ratio

=Energy   stored  in  capacitorWorkdone  by  the  battery=12CV2Ce2= {{Energy\,\,\,stored\,\,in\,\,capacitor} \over {Workdone\,\,by\,\,the\,\,battery}} = {{{1 \over 2}C{V^2}} \over {C{e^2}}}

where

C=C=

Capacitance of capacitor

V=V=

Potential difference,

e=e=
emfemf

of battery

=12Ce2Ce2=12= {{{1 \over 2}C{e^2}} \over {C{e^2}}} = {1 \over 2}
( \left( \, \right.

as

V=eV=e
 )\left. \, \right)
Q62
If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :
A qb=qf(1−1k){q_b} = {q_f}\left( {1 - {1 \over {\sqrt k }}} \right)
B qb=qf(1−1k){q_b} = {q_f}\left( {1 - {1 \over k}} \right)
C qb=qf(1+1k){q_b} = {q_f}\left( {1 + {1 \over {\sqrt k }}} \right)
D qb=qf(1+1k){q_b} = {q_f}\left( {1 + {1 \over k}} \right)
Correct Answer
Option B
Solution

When a dielectric is inserted in a capacitor Due to free charge

E→=E→0\overrightarrow E = {\overrightarrow E _0}

only After dielectric

E′=E0kE' = {{{E_0}} \over k}
qB=qf(1−1k){q_B} = {q_f}\left( {1 - {1 \over k}} \right)
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