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
(ii) V/2
(iii) V/K
(iv) KV
(iii) V/K
Explanation:
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 V/K .
Q = Charge remains constant
Cʹ = K C
Qʹ = Cʹ Vʹ
Q = Cʹ Vʹ
Q = K C Vʹ
Vʹ = `\frac{"Q"}{"KC"}`
= `\frac{"V"}{"K"}`
0 Comments