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A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?
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- The energy stored in the capacitor decreases K times.
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The chance in energy stored is 1 CV² 1 - 1 K - The charge on the capacitor is not conserved.
- The potential difference between the plates decreases K times.
Correct Option: C
Capacitance of the capacitor, C = Q/V
After inserting the dielectric, new capacitance C1 = K.C
New potential difference V1 = V/K
Ui = | CV² = | (∵ Q = CV) | ||
2 | 2C |
Uf = | = | = | = | ![]() | ![]() | ||||
2f | 2KC | 2KC | K |
∆u = uf – ui = | CV² | ![]() | - 1 | ![]() | ||
2 | K |
As the capacitor is isolated, so change will remain conserved p.d. between two plates of the capacitor
L = | = | ||
KC | K |