Question

A parallel air capacitance 245muF has a charge of magnitude 0.48muC on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the surface charge density on each plate?

Answer 1

Given:

Capacitance ($(C)=$  $245 \mu F$

Charge$(Q)$ $=0.48 \mu C$

Distance between plates$(d)$ $=0.328 mm$

To find:  

(a) Potential difference between the plates$(V)$

(b)  Area of each plate$(A)$

(c)  Surface charge density on each plate$(\sigma)$

Solution:

(a) Potential difference between the plates

We know that,

                 $Q=CV$

                  $V = \frac{Q}{C}$

                  $V= \frac{0.48 \mu C}{245 \mu C}$  

                  $V = 1.9 mV$

Thus, Potential difference between the plates is  $1.9mV$

(b)  Area of each plate$(A)$

                             $C = \frac{\epsilon_0 A}{d}$

                              $A = \frac{Cd}{\epsilon_0}$

                             $A = \frac{245 \mu \times 0.328mm}{8.85 \times 10^12}$

                            $A= 9.08 \times 10^3 m^2$

Thus, Area of each plate is $(A) = (9.08 \times 10^3) m^2$

(c)  Surface charge density on each plate$(\sigma)$

                                $\sigma = \frac{Q}{A}$

                                $\sigma = \frac{0.48 \mu C}{9.08 \times 10^8}$

                                $\sigma = 0.052 \times 10^{-6} C/m^2$

Thus,  Surface charge density on each plate is   $0.052 \times 10^{-6} C/m^2$