Electrostatic Energy and Energy Densities of Different Capacitors | Quantum Science Philippines

## Electrostatic Energy and Energy Densities of Different Capacitors

Electrostatic Energy and Energy Densities of Different Capacitors

Author: Quennie J. Paylaga, Master of Science in Physics student

Problem 1.8 (Chapter 1 of Classical Electrodynamics 3rd Edition by JD Jackson)

Calculate the electrostatic energy (express it in terms of equal and opposite charges Q and -Q placed on the conductors and the potential difference between them) and the energy densities of the electrostatic field of the following capacitors:

(a) two large, flat, conducting sheets of area A, separated by a small distance d;

(b) two concentric conducting spheres with radii a, b (b>a);

(c) two concentric conducting cylinders of length L, large compared to their radii a, b (b>a).

Solution:

Before we can solve for the electrostatic energy and energy densities of the three capacitors, we need to solve first the electric field, potential difference and capacitance of each capacitor. The following formula are the tools in solving this problem.

Electric field E (Gauss Law):

$\oint \vec{E} \bullet d \vec{a} = \frac{Q_{enc}}{\epsilon_{o}}$

Potential Difference V:

$V = \int \vec{E} \bullet d \vec{l}$

Capacitance C:

$C = Q / V$

Electrostatic Energy W:

$W = \frac{1}{2} C V^2$

Energy Density w:

$w = \frac{\epsilon_o}{2} \vert E \vert^2$

For Capacitor A

Its electric field is

$\int \vec{E} \bullet d \vec{a} = 2 A \vert \vec{E} \vert$

From Gauss’s Law

$2 A \vert \vec{E} \vert = \frac{Q_{enc}}{\epsilon_{o}} = \frac{\sigma A}{\epsilon_{o}}$

$\vec{E} = \frac{\sigma A}{2 A \epsilon_{o}} = \frac{\sigma}{2 \epsilon_{o}}$

The electric field between the two plates of Capacitor A is

$E = \frac{\sigma}{\epsilon_{o}}$

where $\sigma = \frac{Q}{A}$ (charge per unit length of the parallel-plate capacitor).

The potential difference of Capacitor A is

$V = \int_0^d \vec{E} \bullet d \vec{l} = \int_0^d \frac{\sigma}{\epsilon_{o}} \bullet d \vec{l} = \frac{\sigma d}{\epsilon_{o}} = \frac{Qd}{\epsilon_{o} A}$

The capacitance of Capacitor A is

$C = \frac{Q}{V} = \frac{Q}{\frac{Qd}{\epsilon_{o} A}} = \frac{A \epsilon_{o}}{d}$

Thus, the electrostatic energy of  Capacitor A is

$W = \frac{1}{2} C V^2 = \frac{1}{2} \left(\frac{A \epsilon_o}{d}\right) \left(\frac{Qd}{\epsilon_o A}\right)^2 = \frac{1}{2} \frac{Q^2 d}{\epsilon_o A}$

and its energy density is

$w = \frac{\epsilon_o}{2} \vert E \vert^2 = \frac{\epsilon_o}{2} \left(\frac{\sigma}{\epsilon_o}\right)^2 = \frac{\sigma^2}{2 \epsilon_o} \hspace{0.5cm} for \hspace{0.3cm} 0 < r < d$.

For Capacitor B

The electric field for two conducting spheres is given by:

$\vec{E} = \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2} \hat{r}$

$E = \frac{Q}{4 \pi \epsilon_o r^2}$

Its potential difference is

$V = - \int_b^a \vec{E} \bullet d \vec{l} = - \frac{Q}{4 \pi \epsilon_o} \int_b^a \frac{1}{r^2} dr$

$V = \frac{Q}{4 \pi \epsilon_o} \left(\frac{1}{a} - \frac{1}{b}\right)$

And its capacitance is

$C = \frac{Q}{V} = \frac{Q}{\frac{Q}{4 \pi \epsilon_o} \left(\frac{1}{a} - \frac{1}{b}\right)} = 4 \pi \epsilon_o \frac{ab}{b - a}$

Its electrostatic energy is calculated as

$W = \frac{1}{2} C V^2 = \frac{1}{2} \left(4 \pi \epsilon_o \frac{ab}{b - a}\right) \left(\frac{Q}{4 \pi \epsilon_o} \left(\frac{1}{a} - \frac{1}{b}\right)\right)^2 = \frac{Q^2}{8 \pi \epsilon_o} \left(\frac{b - a}{ab}\right)$

and its energy density is given by

$w = \frac{\epsilon_o}{2} \vert E \vert^2 = \frac{\epsilon_o}{2} \left(\frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}\right)^2 = \frac{Q^2}{32 \pi^2 \epsilon_o} \frac{1}{r^4} \hspace{0.5cm} for \hspace{0.2cm} a

For Capacitor C

The electric field for cylindrical capacitor is

$\vert \vec{E} \vert 2 \pi r L = \frac{Q}{\epsilon_o}$

$E = \frac{Q}{2 \pi r L \epsilon_o}$

Its potential difference is

$V = - \int_b^a \vec{E} \bullet d \vec{l} = - \frac{Q}{2 \pi L \epsilon_o} \int_b^a \frac{1}{r} d r = \frac{Q}{2 \pi L \epsilon_o} ln \frac{b}{a}$

Its capacitance C is

$C = \frac{Q}{V} = \frac{Q}{\frac{Q}{2 \pi L \epsilon_o} ln \frac{b}{a}} = \frac{2 \pi L \epsilon_o}{ln \frac{b}{a}}$

And its electrostatic energy is given by

$W = \frac{1}{2} C V^2 = \frac{1}{2} \left(\frac{2 \pi L \epsilon_o}{ln \frac{b}{a}}\right) \left(\frac{Q}{2 \pi L \epsilon_o} ln \frac{b}{a}\right)^2 = \frac{Q^2}{4 \pi L \epsilon_o} ln \frac{b}{a}$

and its energy density is

$w = \frac{\epsilon_o}{2} \vert E \vert^2 = \frac{\epsilon_o}{2} \left(\frac{Q}{2 \pi L \epsilon_o} \frac{1}{r}\right)^2 = \frac{Q^2}{8 \pi^2 L^2 \epsilon_o} \frac{1}{r^2} = \frac{\lambda^2}{8 \pi^2 \epsilon_o} \frac{1}{r^2} \hspace{0.5cm} for \hspace{0.2cm} a

where $\lambda = \frac{Q}{L}$ (charge per unit length of a cylindrical capacitor).

For the parallel-plate capacitor, the energy density is constant and for the spherical capacitor, the energy is more strongly concentrated close to the inner conductor than in the case of a parallel-cylinder capacitor.