Question

A 10 nano Farad parallel plate capacitor hold the charge of magnitude 50 microcoulomb on each plate if the plates are separated by distance of 0.885 M what is the area of each plate

Answer 1

As per the question, the capacity of the parallel plate capacitor is given as -

              C= 10 nano farad.

                 = $10\times 10^{-9}\ F$

                 = $10^{-8}F$

The separation distance is given as d = 0.885 m.

The charge on each plate q = 50 micro coulomb

                                               = $50\times 10^{-6}\ C$

                                               = $5\times 10^{-5} C$

We know that the permittivity $[\epsilon_{0}]=\ 8.85\times 10^{-12}\ C^2 / Nm^2$

The capacity of a capacitor is calculated as -

                                         $C=\ \frac{\epsilon_{0}A} {d}$

                                         $A=\frac{Cd}{\epsilon_{0}}$

                                         $A=\frac{10^{-8}\times 0.885} {8.85\times 10^{-12}}\ m^2$

                                               $=0.1\times 10^{4} m^2$

                                               $=\ 1.0\times 10^{3}\ m^2$    [ans]

Hence, the area of each plate is $1.0\times 10^3\ m^2$.