Question

How much energy is stored by the electric field between two square plates, 6.1 cm on a side, separated by a 3.5 mm air gap? The charges on the plates are equal and opposite and of magnitude 530 μC. Express your answer using two significant figures

Answer 1

Answer:

$7.6\times 10^{-6} J$

Explanation:

The electric field between two plates is:

$E = \frac{\sigma}{2\epsilon}$

Where sigma is the superficial charge density and epsilon is the permittivity of air (is about the same as vaccum)

$\sigma = \frac{Q}{Area}= \frac{530 \times 10^{-6} C}{(0.061m)^2} = 0.14C/m^2$

Using the formula to determine the electric field.

$E = \frac{0.14C/m^2}{2\times 8.84 \times 10^{-12}C^2N^{-1}m^{-2}}\\E = 8.06 \times 10^9 N/C$

The energy inside the plates is given by the energy density due to electric Field times the volume:

$Energy = \rho_{E} \times V = \frac{1}{2}\epsilon E V$

The volume is equal to the area of the plates times the distance between them:

$Energy = \frac{1}{2}8.82\times 10^{-12}C^2 N^{-1}m^{-2} \times 8.06 \times 10^9 N/C \times (0.061m)^2\times 0.0035m\\Energy = 7.6 \times 10^{-6} J$