Q23CQ - College Physics (Urone)

Question

What happens to the energy stored in a capacitor connected to a battery when a dielectric is inserted? Was work done in the process?

Step-by-Step Solution

Answer

The energy increases because the battery does work on the system Since, energy is increasing, so work is being done.

Step 1: Potential Energy and Capacitance

The electric potential energy U_E stored in a capacitor of capacitance C charged to a potential difference $\mathbf{∆}\mathbf{V}$  is –

 $\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}{\mathbf{\left(}\mathbf{\Delta V}\mathbf{\right)}}^{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(1)}$

The capacitance   of a capacitor filled with a material of dielectric constant k is

 $\mathbf{C}\mathbf{=}{\mathbf{kC}}_{\mathbf{o}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(2)}$

where ${\mathbf{C}}_{\mathbf{°}}$ is the capacitance of the capacitor without a dielectric.

Step 2: Equation for energy stored in a capacitor

The energy stored in the parallel-plate capacitor with no dielectric material is found from Equation ( 1 )  –

 ${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}_{\mathbf{0}}\mathbf{(}\mathbf{\Delta V}{\mathbf{)}}^{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(3)}$

Where ${\mathrm{C}}_{°}$ is capacitance of the capacitor without a dielectric.

Similarly, the energy stored in the capacitor after a dielectric is inserted into the capacitor is –

 $\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{C}\mathbf{(}\mathbf{\Delta V}{\mathbf{)}}^{\mathbf{2}}$

Step 3: Energy rate calculation

Since the capacitor is connected to a battery, the voltage $∆\mathrm{V}$  across the capacitor necessarily remains constant Substituting value for   from Equation ( 2 )  –  

 $\begin{array}{rcl}\mathbf{E}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kC}}_{\mathbf{0}}\mathbf{(}\mathbf{\Delta V}{\mathbf{)}}^{\mathbf{2}}\\ & \mathbf{=}& \mathbf{k}\left[\frac{1}{2}{C}_0{\left(\mathrm{\Delta V}\right)}^2\right]\end{array}$

Substituting for the term in square brackets from Equation ( 3 )  

 $\mathbf{E}\mathbf{=}{\mathbf{kE}}_{\mathbf{0}}$

Since k>1 , the electric potential energy stored in the increases. 

Therefore, as the battery works on the system by pumping charge onto the plates to maintain the potential difference, the energy increases.