Question

A parallel-plate capacitor having plate area 20 cm² and separation between the plates 1⋅00 mm is connected to a battery of 12⋅0 V. The plates are pulled apart to increase the separation to 2⋅0 mm. (a) Calculate the charge flown through the circuit during the process. (b) How much energy is absorbed by the battery during the process? (c) Calculate the stored energy in the electric field before and after the process. (d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart. (e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

Answer 1

Given:
$$\begin{gathered} \mathrm{Area},\mathit{}A=20{\mathrm{cm}}^2=2\times {10}^{-3}{\mathrm{m}}^2 \\ \mathrm{Separation},d_1=1\mathrm{mm}={10}^{-3}\mathrm{m} \end{gathered}$$

Initial capacitance of the capacitor:
$$\begin{gathered} C_1=\frac{{\in }_{0}A}{d_1} \\ {C}_1=\frac{8.88\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}} \\ {C}_1=1.776\times {10}^{-11}\mathrm{F} \end{gathered}$$

Final capacitance of the capacitor:

$C_2=\frac{C_1}{2}\left(\mathrm{because}{d}_2=\frac{d_1}{2}\right)$

(a) Charge flown through the circuit:

$$\begin{gathered} Q=C_{1}V-{\mathrm{C}}_{2}V \\ \mathrm{Q}=\left(C_1-C_2\right)V \\ Q=\frac{1.776}{2}\times {10}^{-11}\times 12.0 \\ Q=1.06\times {10}^{-10}\mathrm{C} \end{gathered}$$

(b) Energy absorbed by the battery:

$$\begin{gathered} E=QV=1.06\times {10}^{-10}\times 12 \\ E=12.72\times {10}^{-10}\mathrm{J} \end{gathered}$$

(c) Energy stored before the process, E_i$=\frac{1}{2}{C}_1{V}^2$
$$\begin{gathered} \Rightarrow E_{\mathrm{i}}=\frac{1}{2}\times 1.776\times {10}^{-11}\times {\left(12\right)}^2 \\ \Rightarrow E_{\mathrm{i}}=12.7\times {10}^{-10}\mathrm{J} \end{gathered}$$

Energy stored after the process, E_f$=\frac{1}{2}{C}_2{V}^2$
$$\begin{gathered} \Rightarrow E_{\mathrm{f}}=\frac{1}{2}\times \frac{1.776}{2}\times {10}^{-11}\times {\left(12\right)}^2 \\ \Rightarrow E_{\mathrm{f}}=6.35\times {10}^{-10}\mathrm{J} \end{gathered}$$

(d) Work done, W = Force × Distance
W = Change in the energy
$\Rightarrow$W = 6.35 × 10⁻¹⁰ J