Electrostatics • Topic 2 of 3

Gauss's Law & Electric Potential

Gauss's law relates the electric flux through any closed surface (a Gaussian surface) to the charge enclosed by it:

$$\oint \vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}$$ It holds for any closed surface and any charge distribution, but it becomes a powerful tool for finding fields only when the symmetry lets us take $E$ out of the integral.

Three standard applications follow directly:

  • Infinite line charge of linear density $\lambda$: $E=\frac{\lambda}{2\pi\epsilon_0 r}$ (field falls as $1/r$, directed radially).
  • Infinite charged sheet of surface density $\sigma$: $E=\frac{\sigma}{2\epsilon_0}$ (uniform, independent of distance).
  • Charged spherical shell of charge $Q$: outside, $E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$ (as if all charge were at the centre); inside, $E=0$.

The electric potential $V$ at a point is the work done per unit positive charge in bringing it from infinity to that point: $V=\frac{W}{q}$, measured in volts (V = J/C). For a point charge,

$$V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$$ Potential is a scalar, so potentials due to many charges simply add algebraically. The potential of a dipole at distance $r$ (for $r\gg a$) is $V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$ — zero on the equatorial line ($\theta=90^\circ$).

Equipotential surfaces are surfaces on which $V$ is constant. No work is done in moving a charge along them, so the electric field is always perpendicular to an equipotential surface. For a point charge they are concentric spheres; for a uniform field they are parallel planes.

Field and potential are linked by $$E=-\frac{dV}{dr}$$ The field points in the direction of decreasing potential, and its magnitude equals the potential gradient. The potential energy of a system of two charges is $U=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}$; for more charges, sum over every distinct pair. A positive $U$ means work was done to assemble the (repelling) system; a negative $U$ means the configuration is bound.

Gaussian surface around a point charge+qGaussian surfaceflux = q / e0
1
Worked Example
A charge of $8.85\times10^{-12}\ \text{C}$ is enclosed by a closed surface. Find the total electric flux through it.
Solution
  1. By Gauss's law $\phi=\frac{q_{enc}}{\epsilon_0}$.
  2. $\phi=\frac{8.85\times10^{-12}}{8.85\times10^{-12}}$.
  3. $\phi=1\ \text{N}\,\text{m}^2\,\text{C}^{-1}$.

Answer: $\phi=1\ \text{N}\,\text{m}^2\,\text{C}^{-1}$.

2
Worked Example
Find the electric field 0.1 m from an infinite line charge with $\lambda=2\times10^{-6}\ \text{C/m}$.
Solution
  1. Use $E=\frac{\lambda}{2\pi\epsilon_0 r}=\frac{2\times(9\times10^{9})\lambda}{r}$.
  2. $E=\frac{2\times9\times10^{9}\times2\times10^{-6}}{0.1}$.
  3. $=\frac{0.036}{0.1}=3.6\times10^{5}\ \text{N/C}$.

Answer: $E=3.6\times10^{5}\ \text{N/C}$.

3
Worked Example
Calculate the potential at a point 0.2 m from a charge of $+4\ \mu\text{C}$.
Solution
  1. Use $V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$.
  2. $V=9\times10^{9}\times\frac{4\times10^{-6}}{0.2}$.
  3. $=9\times10^{9}\times2\times10^{-5}=1.8\times10^{5}\ \text{V}$.

Answer: $V=1.8\times10^{5}\ \text{V}$.

4
Worked Example
The potential varies as $V=6x^2$ (in volts, $x$ in metres). Find the electric field at $x=2\ \text{m}$.
Solution
  1. Use $E=-\frac{dV}{dx}$.
  2. $\frac{dV}{dx}=12x$, so $E=-12x$.
  3. At $x=2$: $E=-24\ \text{V/m}$ (i.e. magnitude $24\ \text{V/m}$ in the $-x$ direction).

Answer: $E=-24\ \text{V/m}$ (24 V/m pointing toward decreasing $x$).

5
Worked Example
Two charges $+3\ \mu\text{C}$ and $-3\ \mu\text{C}$ are 0.6 m apart. Find the potential energy of the system.
Solution
  1. Use $U=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}$.
  2. $U=9\times10^{9}\times\frac{(3\times10^{-6})(-3\times10^{-6})}{0.6}$.
  3. $=9\times10^{9}\times\frac{-9\times10^{-12}}{0.6}=\frac{-0.081}{0.6}$.
  4. $U=-0.135\ \text{J}$.

Answer: $U=-0.135\ \text{J}$ (bound system).

6
Worked Example
An infinite sheet has surface charge density $\sigma=4\times10^{-7}\ \text{C/m}^2$. Find the field just outside it.
Solution
  1. Use $E=\frac{\sigma}{2\epsilon_0}$.
  2. $E=\frac{4\times10^{-7}}{2\times8.85\times10^{-12}}$.
  3. $=\frac{4\times10^{-7}}{1.77\times10^{-11}}$.
  4. $E\approx2.26\times10^{4}\ \text{N/C}$.

Answer: $E\approx2.26\times10^{4}\ \text{N/C}$.

Key Points

  • Gauss's law: $\oint\vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}$; total flux depends only on the enclosed charge.
  • Standard fields: line $E=\frac{\lambda}{2\pi\epsilon_0 r}$, sheet $E=\frac{\sigma}{2\epsilon_0}$, shell $E=0$ inside and $\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$ outside.
  • Potential of a point charge $V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$ is a scalar; potentials add algebraically. Dipole: $V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$.
  • Field is perpendicular to equipotential surfaces and $E=-\frac{dV}{dr}$ — it points toward decreasing potential.
  • Potential energy of two charges $U=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}$; negative $U$ means a bound (attractive) configuration.
Tap an option to check your answer0 / 4
Q1.Gauss's law states that the total flux through a closed surface equals:
Explanation: By Gauss's law $\oint\vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}$.
Q2.The electric field inside a uniformly charged conducting shell is:
Explanation: No charge is enclosed by a Gaussian surface inside the shell, so $E=0$.
Q3.Electric potential is a:
Explanation: Potential has magnitude only; contributions from several charges add algebraically.
Q4.The field of an infinite charged sheet:
Explanation: $E=\frac{\sigma}{2\epsilon_0}$ is uniform and independent of distance from the sheet.