Question

Two identified parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Answer 1

Hint: It is given that when the switch is closed, there is a current V flowing through the capacitors. Find the total energy by adding up the energy contributed by each capacitor. When the switch is opened and the di-electric is filled in between, calculate the total energy in this case and find the ratio of the same.

Complete step by step answer
When switch S is closed, there is a current I flowing through the circuit which has a potential difference of V volts from the source. Energy emitted by a capacitor in this case is expressed as the product of its capacitance and the square of the voltage flowing through the capacitor. This is mathematically represented as ,
 $\Rightarrow Q={\displaystyle \frac{1}{2}}C{V}^2$ (where C is the capacitance of the capacitor and V is the voltage flowing across)
Now when switch S is closed, the total energy given out by the system is given as,
 $\Rightarrow E=Q_A+Q_B$
Where $Q_A$ is energy given out by capacitor A and $Q_B$ is energy given out by capacitor B.
 $\Rightarrow Q_A=Q_B={\displaystyle \frac{1}{2}}C{V}^2$
The overall energy can be written as ,
 $\Rightarrow E=C{V}^2$
Now, when the switch is open and the di-electric of di-electric constant K is inserted between the parallel plate capacitors, there will be a flow of voltage V across the capacitor A. Whereas in capacitor B the stored voltage will be flowing across the plates, which is denoted by $V^{\prime }$ . Now , total energy of the system is again the sum of the individual energy possessed by the capacitors
 $\Rightarrow E^{\prime }=Q_A+Q_B$
Now, energy across the capacitor A is given as,
 $\Rightarrow Q_A={\displaystyle \frac{1}{2}}(KC)V^2$ , Where K is the di-electric constant.
In case of capacitor B, the voltage flowing across the plates will be the stored voltage $V^{\prime }$ . This can be represented as $V^{\prime }={\displaystyle \frac{V}{K}}$ . Using this, the energy across capacitor B is,
 $\Rightarrow Q_B={\displaystyle \frac{1}{2}}(KC)V{}^{\prime 2}$
On substituting and cancelling the common term, we get,
 $\Rightarrow Q_B={\displaystyle \frac{1}{2}}{\displaystyle \frac{C{V}^2}{K}}$
Now total energy $E^{\prime }$ is given as,
 $\Rightarrow E^{\prime }=Q_A+Q_B$
On substituting ,
 $\Rightarrow E^{\prime }={\displaystyle \frac{1}{2}}KC{V}^2+{\displaystyle \frac{1}{2}}{\displaystyle \frac{C{V}^2}{K}}$
Taking the common term, we get,
 $\Rightarrow E^{\prime }={\displaystyle \frac{1}{2}}C{V}^2(K+{\displaystyle \frac{1}{K}})$
On simplifying, we get,
 $\Rightarrow E^{\prime }={\displaystyle \frac{1}{2}}C{V}^2({\displaystyle \frac{K^2+1}{K}})$
Now, the ratio between the first and second case is given as ,
 $\Rightarrow {\displaystyle \frac{E}{E^{\prime }}}={\displaystyle \frac{C{V}^2}{{\displaystyle \frac{1}{2}}C{V}^2({\displaystyle \frac{K^2+1}{K}})}}$
On removing the common term and taking the denominator term to numerator , we get,
 $\Rightarrow {\displaystyle \frac{E}{E^{\prime }}}={\displaystyle \frac{2K}{K^2+1}}$
Hence, the ratio of the total electrostatic energy stored in both capacitors before and after introduction of di-electric is found out.

Note
In a parallel plate capacitor setup, the capacitance depends upon the charge flowing between the plate and the Voltage applied through the circuit, the area between the plate and the permissibility factor.