A$\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{\mu F}$ capacitor is connected to a power supply that keeps a constant potential difference of 24.0V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

The energy storage before inserting the dielectric material is${\mathrm{U}}_{\mathrm{o}}$ and it is given by the equation

${\mathbf{U}}_{\mathbf{o}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}_{\mathbf{o}}{\mathbf{V}}^{\mathbf{2}}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(1\right)$

where${\mathrm{C}}_{\mathrm{o}}$ is the capacitance before the dielectric material andVis the potential difference which is constant before and after the dielectric material

Using a dielectric allows a capacitor to sustain a higher potential difference Vbut as the voltage is constant, therefore the capacitor will sustain a higher capacitance, hence higher energy where the new capacitance is given by the equation

role="math" localid="1664254597173" $\mathbf{C}\mathbf{=}{\mathbf{CK}}_{\mathbf{o}}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(2\right)$

where Cis the capacitance with the dielectric material and Kis the dielectric constant and is unitless and higher than unity

The energy storage U after the dielectric material

$\mathrm{U}=\frac{1}{2}{\mathrm{CV}}^2$

using equation $\left(2\right)$

$\mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{KC}}_{\mathbf{o}}{\mathbf{V}}^{\mathbf{2}}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(3\right)$

(a)

We are given a capacitor without a dielectric material with capacitance

$$\begin{gathered} {\mathrm{C}}_{\mathrm{o}}=12.50\mathrm{\mu C} \\ =12.50\times {10}^{-6}\mathrm{C} \\ \mathrm{V}=24.0\mathrm{V} \end{gathered}$$

the dielectric constant for the dielectric material is $\mathrm{K}=3.75$

Now, using equation (1)

The energy storage of the capacitor before the dielectric material

$$\begin{gathered} {\mathrm{U}}_{\mathrm{o}}=\frac{1}{2}{\mathrm{C}}_{\mathrm{o}}{\mathrm{V}}^2 \\ =\frac{1}{2}\left(12.50\times {10}^{-6}\mathrm{C}\right){\left(24.0\mathrm{V}\right)}^2 \\ =0.0036\mathrm{J} \end{gathered}$$

Hence, the energy storage of the capacitor before the dielectric material is $0.0036\mathrm{J}$

Using equation (2)

the energy storageafter the dielectric material

$$\begin{gathered} \mathrm{U}=\frac{1}{2}{\mathrm{KC}}_{\mathrm{o}}{\mathrm{V}}^2 \\ =\frac{1}{2}\left(3.75\right)\left(12.50\times {10}^6\mathrm{C}\right){\left(24.0\mathrm{V}\right)}^2 \\ =0.0135\mathrm{J} \end{gathered}$$

Hence, the energy storage after the dielectric material is.

(b)

The change in energy

$$\begin{gathered} ∆\mathrm{U}=\mathrm{U}-{\mathrm{U}}_{\mathrm{o}} \\ =0.0135\mathrm{J}-0.0036\mathrm{J} \\ =0.0099\mathrm{J} \end{gathered}$$

Therefore, the energy increases during the insertion of the dielectric material.