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 …

Question

asked Jan 12, 2020 178k views

1 Answer

Yftach · Answer 1 · 2020-01-18T05:55:34+0000

Answer:

7.6* 10^(-6) J

Step-by-step explanation:

The electric field between two plates is:

$E = (\sigma)/(2\epsilon)$

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

$\sigma = (Q)/(Area)= (530 * 10^(-6) C)/((0.061m)^2) = 0.14C/m^2$

Using the formula to determine the electric field.

$E = (0.14C/m^2)/(2* 8.84 * 10^(-12)C^2N^(-1)m^(-2))\\E = 8.06 * 10^9 N/C$

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

$Energy = \rho_(E) * V = (1)/(2)\epsilon E V\\$

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

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