Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant K. The potential differences across the capacitors now becomes...........

Options

A. `(3V)/(K+2)`
B. `(3V)/(K)`
C. `(V)/(K+2)`
D. `(V)/(K)`

Answer

A

Explanation

The condensers are joined in parallel. Hence they have the same P.D. while charging
`therefore Q_(1)=(2C)V and Q_(2)=CV`
and `Q_(1)+Q_(2)=3CV`
After removing the battery, the condenser of capacity C is filled with a dielectric. Hence its capacity will be WK. `therefore`The resultant capacity`=2C+KC=C(2+K)` . . (2)
The total charge will remain the same i.e. `Q_(1)+Q_(2)`
But `because V=(Q)/(C)`, V will change to V'
`therefore Q_(1)+Q_(2)=C(2+K)V' " "therefore 3CV=C(2+K)V'`
`therefore V'=(3V)/(K+2)`