Question

Two large, non-conducting plates with the indicated area charge densities are suspended 7.79 cm apart. You may treat each plate as an infinite sheet. How much electrostatic energy is stored in 4.54 cubic centimeter of the space in region A (left of both plates)?

What volume of the space in region B (between the plates) stores an equal amount of energy?

The plates are +81.3 uC/m^2 and +23.6 uC/m^2.

Answer 1

a)

Electric field in region A is

$E_{A}=-\frac{\sigma _{1}}{2\varepsilon _{o}}-\frac{\sigma _{2}}{2\varepsilon _{o}}=-\frac{1}{2\times (8.8542\times 10^{-12})}(81.3+23.6)\times 10^{-6}$

E_A=-5.9237*10⁶ N/C

Electrostatic energy stored in region A is

$U=\frac{1}{2}\varepsilon _{o}E^{2}V=\frac{1}{2}\times (8.8542\times 10^{-12})(-5.9237\times 10^{6})^{2}\times (4.54\times 10^{-6})$

U=7.053*10^-4 J

b)

Electric field in region B is

$E_{B}=\frac{\sigma _{1}}{2\varepsilon _{o}}-\frac{\sigma _{2}}{2\varepsilon _{o}}=-\frac{1}{2\times (8.8542\times 10^{-12})}(81.3-23.6)\times 10^{-6}$

E_B=3.2583*10⁶N/C

Since

$U=\frac{1}{2}\varepsilon _{o}E_{B}^{2}V_{B}=\frac{1}{2}\times (8.8542\times 10^{-12})(3.2583\times 10^{6})^{2}\times V_{B}$

$7.054\times 10^{-4}=\frac{1}{2}\times (8.8542\times 10^{-12})(3.2583\times 10^{6})^{2}\times V_{B}$

V_B=1.5*10^-5 m³=15 cm³