A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,
electric intensity incrases
potential increases
energy decreases
capacity decreases
A
electric intensity incrases
B
energy decreases
C
potential increases
D
capacity decreases
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Solution
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→ When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge) → We know C=ε0Ad When slab is introduced, C=kε0Ad C increases and Q is constant. We know: Q=CV
∴V has to decrease. We know energy store in capacitor =12QV ∴ as V decreases, energy decreases.
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