The plates of a parallel plate capacitor in vacuum are 5.0 mm apart. A potential difference of 10.0 kV is applied. The magnitude of the surface charge density on each plate is
Answer & explanation
Correct answer: option 1
$ \text{ Charge on plates of capacitor is } Q = CV , C =\frac{\epsilon_0 A}{d}$
$ \text{Surface Charge density } \sigma = \frac{Q}{A} = \frac{CV}{A} = \frac{\epsilon_0 V}{d}$
$ \sigma = \frac{8.85\times 10^{-12}\times 10^4}{5\times 10^{-3}} = 1.77 \times 10^{-5} C/m^2$