Electrostatic Potential and Capacitance - Result Question 28
30. A parallel plate air capacitor has capacity ${ }^{\prime} C$ ’ distance of separation between plates is ’ $d$ ’ and potential difference ’ $V$ ’ is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is: [2015 RS]
(a) $\frac{CV^{2}}{2 d}$
(b) $\frac{CV^{2}}{d}$
(c) $\frac{C^{2} V^{2}}{2 d^{2}}$
(d) $\frac{C^{2} V^{2}}{2 d^{2}}$
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Answer:
Correct Answer: 30. (a)
Solution:
- (a) Force of attraction between the plates, $F=qE$
$=q \times \frac{\sigma}{2 \epsilon_0}$
$=q \frac{q}{2 A \epsilon_0} \quad[\because E=\frac{6}{2 E_o}]$
$=\frac{q^{2}}{2(\frac{\epsilon_0 A}{d}) \times d}=\frac{C^{2} V^{2}}{2 C d}=\frac{C V^{2}}{2 d}$
Here, $C=\frac{\epsilon_0 A}{d}, q=CV, A=$ area
To find the force of attraction between the plates, we use the concept that work done in displacing the plates against the force is equal to the increase in energy of the capacitor.
