A parallel plate capacitor has a potential of 100 V and is disconnected from the
charging battery. A dielectric material of constant K = 10 is put between its
plates. The new potential will be
Options:
1000 V
110 V
10 V
90 V
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2
Answer:
The capacitance without dielectric is C=
d
Aϵ
0
When dielectric slab is inserted, the capacitance becomes, C
′
=
d
AKϵ
0
=KC where K be the dielectric constant.
i)Thus, the capacitance will increase K times of the initial.
ii) As the battery is disconnected so the charge on capacitor remains constant.
Since, Q=CV so potential V will decrease and also E=V/d so the field E will also decrease.
iii) Stored energy, U=
2C
Q
2
. As charge Q is constant and C is increasing so energy will decrease.
Explanation:
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