Question

A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed $v$. If $ x $ is the separation between the plates, then the time rate of change of the electrostatic energy of the condenser is proportional to:

• A. $ {{x}^{2}} $
• B. $ x $
• C. $ \frac{1}{x} $
• D. $ \frac{1}{{{x}^{2}}} $

Answer 1

Answer: (D)

$U=\frac{1}{2} C V^{2}=\frac{1}{2} \frac{\varepsilon_{0} A}{d} V^{2}$
At any instant, let the separation between plates be$x$ .
So, $ U=\frac{1}{2} \frac{\varepsilon_{0} A}{x} V^{2}$
$\therefore \frac{d U}{d t}=\frac{1}{2} \varepsilon_{0} A V^{2}(-1) \frac{1}{x^{2}}=\frac{d x}{d t} $
$=-\frac{1}{2} \frac{\varepsilon_{0} A V^{2}}{x^{2}}(v)$
i.e., potential energy decreases as $\left(1 / x^{2}\right)$.