Question

A large capacitor in an air conditioner is typically connected to the potential difference provided to your house (120 V). Although capacitors like this are usually cylindrical, model this one as a parallel plate capacitor with square plates separated by 2.0 mm. If you want your capacitor to store 0.25 J of energy, what is the area of the capacitor plates?

Answer 1

Answer:

$A=7.72\times 10^{-18}\ m^2$

Explanation:

Given:

• the voltage of the source, $V=120\ V$

• distance of separation between the capacitor plates, $d=2\times 10^{-3}\ J$

• energy of the capacitor, $U=0.25\ J$

Assuming that there is free space between the plates, so the permittivity, $\epsilon=9\times 10^{9}\ N.m^2.C^{-2}$

We have the energy stored in a capacitor as:

$U=\frac{1}{2}\times \frac{\epsilon.A.V^2}{d}$

$0.25=0.5\times \frac{9\times 10^{9}\times A\times 120^2}{2\times 10^{-3}}$

$A=7.72\times 10^{-18}\ m^2$

Answer 2

Explanation:

Formula for energy stored in the capacitor is as follows.

               E = $\frac{1}{2}CV^{2}$

or,          C = $\frac{2E}{V^{2}}$

The given data is as follows.

                  E = 0.25 J,        V = 120 V

Putting the given values into the above formula as follows.

                   C = $\frac{2E}{V^{2}}$

                       = $\frac{2 \times 0.25 J}{(120)^{2}}$

                       = $3.47 \times 10^{-5}$ F

Also,          C = $\frac{\epsilon_{o} A}{d}$

or,               A = $\frac{Cd}{\epsilon_{o}}$

Putting the given values into the above formula as follows.

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

                  = $\frac{3.47 \times 10^{-5} \times 2 \times 10^{-3}}{8.85 \times 10^{-12}}$

                  = 7846.82 $m^{2}$

Thus, we can conclude that area of the capacitor plates is 7846.82 $m^{2}$.