Question

Question: (II) How much energy is stored by the electric field between two square plates, 8.0 cm on a side, separated by a 1.5-mm air gap? The charges on the plates are equal and opposite and of magnitude $\mathbf{370}\mu \mathbf{C}$.

Short answer

The energy stored by the electric field between two square plates is $1.8\times {10}^3\mathrm{J}$.

Step-by-step solution

Understanding the energy stored in a capacitor

A capacitor is a charge and energy storage device. The stored energy is proportional to the square of the magnitude of the charge on the plates.

The expression for stored energy in the capacitor is given as:

$\mathrm{P}\mathrm{E}=\frac{Q^2}{2C}$ … (i)

Here, C is the capacitance and Q is the charge on the capacitor plates.

Given data

The side of the square plates is,$s=8.0\mathrm{c}\mathrm{m}=0.08\mathrm{m}$.

The separation of the plates is,$d=1.5\mathrm{m}\mathrm{m}=1.5\times {10}^{-3}\mathrm{m}$.

The magnitude of the charge on the plates is, $Q=370\mu \mathrm{C}=370\times {10}^{-6}\mathrm{C}$.

Determination of the capacitance

The area of the square plate is,

$A=s^2$

The capacitance of the capacitor is given as:

$\begin{array}{r}lC={\epsilon }_0\frac{A}{d}\\ C={\epsilon }_0\frac{s^2}{d}\end{array}$

Substitute the values in the above expression.

$\begin{array}{rl}cC& =\left(8.854\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2\right)\frac{{\left(0.08\mathrm{m}\right)}^2}{\left(1.5\times {10}^{-3}\mathrm{m}\right)}\\ C& =3.78\times {10}^{-11}\mathrm{F}\end{array}$

Determination of the energy stored in the capacitor

The energy stored in the capacitor is given as:

$\mathrm{P}\mathrm{E}=\frac{Q^2}{2C}$

Substitute the values in the above expression.

$\begin{array}{rl}c\mathrm{P}\mathrm{E}& =\frac{{\left(370\times {10}^{-6}\mathrm{C}\right)}^2}{2\times 3.78\times {10}^{-11}\mathrm{F}}\\ & =1.8\times {10}^3\mathrm{J}\end{array}$

Thus, the energy stored by the electric field between two square plates is $1.8\times {10}^3\mathrm{J}$.