91Ó°ÊÓ

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

$$\begin{gathered} {\mathrm{C}}_{\mathrm{o}}=12.50\mathrm{μ°ä} \\ =12.50×{10}^{-6}\mathrm{C} \\ \mathrm{V}=24.0\mathrm{V} \end{gathered}$$

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

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

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

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

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.