A capacitor plates are charged by a battery with ‘V’ volts. After charging battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become
(i) Zero
(ii) V/2
(iii) V/K
(iv) KV
(i) Zero
(ii) V/2
(iii) V/K
(iv) KV
Answers & Comments
Verified answer
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When a dielectric slab with dielectric constant K is inserted between the plates of a capacitor, the potential difference V across the plates does not change. The potential difference remains the same as it was when the plates were charged by the battery.
So, the potential across the plates of the capacitor remains V.
The correct option is:
(iii) V/K
Answer:
Option 3} is the correct answer.
Explanation:
Capacitor plates are charged by a battery with 'V' volts. After charging the battery is disconnected and a dielectric slab with dielectric constant 'K' is inserted between its plates, the potential across the plates of a capacitor will become V/K.