Question

A parallel-plate capacitor having plate area 400 cm² and separation between the plates 1⋅0 mm is connected to a power supply of 100 V. A dielectric slab of thickness 0⋅5 mm and dielectric constant 5⋅0 is inserted into the gap. (a) Find the increase in electrostatic energy. (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?

Answer 1

The capacitance of the capacitor without the dielectric slab is given by

$C_1=\frac{{\in }_{0}A}{d}$

$$\begin{gathered} \Rightarrow C=\frac{(8.85\times {10}^{-12})\times (400\times {10}^{-4})}{(1.0\times {10}^{-3})} \\ \Rightarrow C=3.54\times {10}^{-10}\mathrm{F} \end{gathered}$$

When the dielectric slab is inserted, the capacitance becomes
$$\begin{gathered} C\text{'}=\frac{{\in }_0\mathit{}A}{\left(d-{\displaystyle t+\frac{t}{K}}\right)} \\ \Rightarrow C\text{'}=\frac{(8.85\times {10}^{-12})\times (400\times {10}^{-4})}{\left(1\times {10}^{-3}-0.5\times {10}^{-3}+{\displaystyle \frac{0.5\times {10}^{-3}}{5}}\right)} \\ \Rightarrow C\text{'}=\frac{5\times (8.85\times {10}^{-12})\times (400\times {10}^{-4})}{(6\times 0.5\times {10}^{-3})} \\ \Rightarrow C\text{'}=5.9\times {10}^{-10}\mathrm{F} \end{gathered}$$

(a) Increased electrostatic energy:

$$\begin{gathered} ∆E=\frac{1}{2}C\text{'}{V}^2-\frac{1}{2}\mathrm{C}{V}^2 \\ \Rightarrow ∆E=\frac{1}{2}\left(5.9-3.54\right)\times {10}^{-10}\times {\left(100\right)}^2 \\ \Rightarrow ∆E=1.18\times {10}^{-6}\mathrm{J}=1.18\mathrm{\mu J} \end{gathered}$$

(b)
Charge on the capacitor containing dielectric is,
$$\begin{gathered} Q\text{'}=C\text{'}V \\ \Rightarrow Q\text{'}=5.9\times {10}^{-10}\times 100 \\ \Rightarrow Q\text{'}=5.9\times {10}^{-8}\mathrm{C} \end{gathered}$$

Potential difference across the capacitor after removing battery and then dielectric is,
$$\begin{gathered} V\text{'}=\frac{Q\text{'}}{C} \\ \Rightarrow V\text{'}=\frac{5.9\times {10}^{-8}}{3.54\times {10}^{-10}} \\ \Rightarrow V\text{'}=166\mathrm{V} \end{gathered}$$

Increased electrostatic energy,
$$\begin{gathered} ∆E\text{'}=\frac{1}{2}CV{\text{'}}^2-\frac{1}{2}\mathrm{C}\text{'}{V}^2 \\ \Rightarrow ∆E\text{'}=\frac{1}{2}\left[3.54\times {10}^{-10}\times (166{)}^2-5.9\times {10}^{-10}\times (100{)}^2\right] \\ \Rightarrow ∆E\text{'}=1.92\times {10}^{-6}\mathrm{J}=1.92\mathrm{\mu J} \end{gathered}$$

(c) When the battery is connected, energy is increased after insertion of the dielectric slab because of the increase in the capacitance of the capacitor. Now, capacitor will abstract more change from the battery.

When the battery is disconnected, and dielectric slab is taken out then energy stored in the will increase because of increase of potential difference across the capacitor.