Question

Two 50 mu F capacitors connected in parallel are charged by using a 12.0V battery.As soon as the battery is removed then an insulator of dielectic constant kappa=5.0 is inserted to the filling spaces between these capacitors. Calculate the energy stored in the capacitor system after insertion of insulators between two capacitors.​

Answer 1

Given that,

Capacitors = 50 μF

Potential difference = 12.0 V

We need to calculate the net capacitors

Using formula of parallel

$C=C_{1}+C_{2}$

Put the value into the formula

$C=50+50=100\ \mu F$

We need to calculate the charge

Using formula of potential difference

$V = \dfrac{Q}{C}$

$Q=VC$

Put the value into the formula

$Q=12.0\times100\times10^{-6}$

$Q=12\times10^{-4}\ C$

We need to calculate the energy stored in the capacitor system

Using formula of energy

$U=\dfrac{1}{2}\times\dfrac{Q^2}{C}$

Put the value into the formula

$U=\dfrac{1}{2}\times\dfrac{(12\times10^{-4})^2}{100\times10^{-6}}$

$U=0.0072\ J$

$U=7.2\ mJ$

The capacitance increases as the dielectric constant K>1

The capacitance increases then the energy will decrease.

Hence, The energy stored in the capacitor system after insertion of insulators between two capacitors is 7.2 mJ.