Question

Two parallel plate capacitor of capacitance C and 2C are connected in parallel and charged to potential 24 V. The battery is then disconnected and the region between the plates of capacitor C is completely filled material of dielectric constant 2. The potential (in V) difference across capacitor is now nv. Value of n is

Answer 1

Answer:

thanks jarur dena

Explanation:

Two parallel plate capacitor of capacitance C and 2C are connected in parallel and charged to potential 24 V. The battery is then disconnected and the region between the plates of capacitor C is completely filled material of dielectric constant 2. The potential (in V) difference across capacitor is now nv. Value of n is

ANSWER

Initial total charge of the system is: Q

i

=Q

1

+Q

2

=CV+2CV=3CV

When dielectric is inserted in C so the capacitance becomes KC

Final charge, Q

f

=Q

1

′

+Q

2

′

=KCV

′

+2CV

′

=(K+2)CV

′

where V

′

= common potential after disconnected the battery.

As battery is disconnected so total charge will remain unchanged.

Thus, Q

i

=Q

f

⇒3CV=(K+2)CV

′

∴V

′

=

K+2

3V