Question

The energy required to charge a parallel plate condenser of plate separation $ d $ and plate area of cross-section $ A $ such that the uniform electric field between the plates is $ E $ , is

• A. $ \frac{1}{2}\,\varepsilon_{0}\,E^{2} /Ad $
• B. $ \varepsilon_{0}\,E^{2}/Ad $
• C. $ \varepsilon_{0}\,E^{2}Ad $
• D. $ \frac{1}{2}\,\varepsilon_{0}\,E^{2}Ad $

Answer 1

Answer: (C)

Energy given by the cell
$E=C V^{2}$
Here, $C=$ capacitance of condenser
$=\frac{A \varepsilon_{0}}{d}$
$V=$ potential difference across the plates $=E d$
Therefore, $E=\left(\frac{A \varepsilon_{0}}{d}\right)(E d)^{2}$
$=A \varepsilon_{0} E^{2}d$