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

Options

A. Zero
B. `(V)/(2)`
C. `(V)/(k)`
D. KV

Answer

C

Explanation

Given : Capacitor of capacitance = C
Charging voltage = V
Dielectric constant =K
Case 1: When a capacitor is initially charged
it acquires charge. q = CV

Case 2: When battery is disconnected its charge remains **q**. But when a dielectric slab is introduced within the plates the only parameter that remains constant is the charge **q**.
Hence `.q.=C_(0)V_(0)=CV`
Since `C_(0)=epsi_(0)=(A)/(d)` in case 1 and C=
`K epsi_(0)(A)/(d)` in case 2 . which means C=
`KC_(0)`
`:.q=C_(0)V_(0)=kC_(0)V`
`rArr V.=(C_(0)V_(0))/(KC_(0))=(V_(0))/(K)`