A parallel-plate capacitor having plate area 20cm^2 and separatio… - Vidyadip

Question

A parallel-plate capacitor having plate area $20cm^2$ and separation between the plates 1.00mm is connected to a battery of 12.0V. The plates are pulled apart to increase the separation to 2.0mm.

Calculate the charge flown through the circuit during the process.
How much energy is absorbed by the battery during the process?
Calculate the stored energy in the electric field before and after the process.
Using the expression for the force between the plates, find the work done by the person pulling the plates apart.
Show and justify that no heat is produced during this transfer of charge as the separation is increased.

✓

Answer

Area $= a = 20cm^2 = 2 \times 10^{-2}m^2$
d = Separation $= 1mm = 10^{-3}m$
$\text{C}_{\text{i}}=\frac{\in_0\times2\times10^{-3}}{10^{-3}}=2\in_0$
$\text{C}_{\text{f}}=\frac{\in_0\times2\times10^{-3}}{2\times10^{-3}}=\in_0$
$\text{q}_{\text{i}}=24\in_0$
$\text{q}_{\text{f}}=12\in_0$

So, q flown out $12\in_0.$ i.e., $q_i - q_f$

So, $q = 12 \times 8.85 \times 10^{-12} = 106.2 \times 10^{-12}C = 1.06 \times 10^{-10}C$

Energy absorbed by battery during the process

$= q \times v = 1.06 \times 10^{-10}C \times 12$
$= 12.7 \times 10^{-10}J$

Before the process

$\text{E}_{\text{i}}=\Big(\frac{1}{2}\Big)\times\text{C}_{\text{i}}\times\text{V}^2$
$=\Big(\frac{1}{2}\Big)\times2\times8.85\times10^{-12}\times144$
$=12.7\times10^{-10}\text{J}$
After the force
$\text{E}_{\text{f}}=\Big(\frac{1}{2}\Big)\times\text{C}_{\text{f}}\times\text{V}^2$
$=\Big(\frac{1}{2}\Big)\times8.85\times10^{-12}\times144$
$=6.35\times10^{-10}\text{J}$

Workdone = Force × Distance

$=\frac{1}{2}\frac{\text{q}^2}{\in_0\text{A}}=1\times10^{3}$
$=\frac{1}{2}\times\frac{12\times12\times\in_0\times\in_0\times10^{-3}}{\in_0\times2\times10^{-3}}$

From (c) and (d) we have calculated, the energy loss by the separation of plates is equal to the work done by the man on plate. Hence no heat is produced in transformer.

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