Question : 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
a) Zero
b) V/2
c) V/K
d) KV
Doubt by Tushar
Solution :
We know,
When the battery is disconnected then charge on the capacitor becomes constant so
Q = cosnt.
When the battery is disconnected then charge on the capacitor becomes constant so
Q = cosnt.
When a dielectric slab in inserted between the plates of capacitor then
C = KC₀
Also, as per the relation
Q =CV
V = Q/C
C = KC₀
Also, as per the relation
Q =CV
V = Q/C
V ∝ 1/C [∵ Q = cosnt.]
V = Q/KC₀
V = (Q/C₀) × (1/K)
V = V₀ × (1/K)
V = V₀/K
V = Q/KC₀
V = (Q/C₀) × (1/K)
V = V₀ × (1/K)
V = V₀/K
Hence, when the a capacitor is charged then disconnected from the battery and a dielectric slab inserted between the plates then the potential between the plates reduced by K times.
Hence, c) would be the correct answer.
