Coulomb's Law and Gauss's Law Lecture

Coulomb's Law

• Coulomb's Law describes the electrostatic interaction between electrically charged particles.

• The electric field intensity $E$ at a point $R$ due to a charge $q$ is given by:
  $E = \frac{q}{4 \pi \epsilon_0 R^2} \hat{R} \quad (V/m)$
  where:

  • $\epsilon_0$ is the permittivity of free space.
  • $\hat{R}$ is the unit vector pointing from the charge to the point of interest.

• For a line charge density $\rho<em>l$ along the z-axis, the electric field at a point $P(r, \phi, z)$ is: $E(P) = \int \frac{dq}{4 \pi \epsilon</em>0 R'^2} \hat{R} = \int \frac{\rho<em>l dl'}{4 \pi \epsilon</em>0 R'^2} \hat{R}$
  where:

  • $dq = \rho_l dl'$ is the differential charge element.
  • $R'$ is the distance from the charge element to the point $P$.

• The electric field due to an infinite line charge is:
  $E = \frac{\rho<em>l}{2 \pi \epsilon</em>0 r} \hat{r} \quad (V/m)$
  where:

  • $r$ is the radial distance from the line charge.

Gauss's Law

• Gauss's Law relates the electric flux through a closed surface to the enclosed charge.

• Differential Form:
  $\nabla \cdot D = \rho_v$
  where:

  • $D$ is the electric flux density.
  • $\rho_v$ is the volume charge density.

• Integral Form:
  $\oint<em>S D \cdot ds = Q</em>{encl}$
  where:

  • $S$ is the Gaussian surface enclosing volume $V$.
  • $Q_{encl}$ is the total charge enclosed within the volume $V$.

• Divergence Theorem:
  $\int<em>V (\nabla \cdot D) dV = \oint</em>S D \cdot ds$

• Gauss's law states that the total electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space ($\epsilon<em>0$): $\oint</em>S D \cdot dS = Q<em>{incl} = \int</em>V \rho dV$
  $\oint<em>S \epsilon</em>0 E \cdot dS = Q_{incl}$

• If there is no enclosed charge ($Q<em>{incl} = 0$), then: $\oint</em>S D \cdot dS = 0$

Applying Gauss's Law

• Electric Field due to a Single Charge:

  • Gaussian surface: Sphere of radius $R$.
  • $\oint<em>S E \cdot dS = E \oint</em>S dS = E(4 \pi R^2) = \frac{Q}{\epsilon_0}$

• Electric Field due to an Infinite Line Charge Density:

  • Gaussian surface: Cylinder.
  • $E = \frac{\rho<em>l}{2 \pi \epsilon</em>0 r} \hat{r} \quad (V/m)$

• Example: Cylinder of radius r and height h

  • The flux through the top and bottom surfaces is zero:
    $\int<em>{S</em>{top}} D \cdot dS = \int<em>{S</em>{bottom}} D \cdot dS = 0$
  • The flux through the cylindrical surface is:
    $\oint<em>S \vec{D} \cdot d\vec{S} = D</em>r \oint dS = D_r (2 \pi r h)$
  • Total charge enclosed:
    $Q = \rho_l h$
  • Applying Gauss's Law:
    $D<em>r (2 \pi r h) = \rho</em>l h$
    $D = \frac{\rho<em>l}{2 \pi r} \hat{r}$$E = \frac{\rho</em>l}{2 \pi \epsilon_0 r} \hat{r}$

2016 Test 2 Q1

• Problem: A cylindrical volume with radius $r = a$ contains a charge density given by $\rho<em>v = \rho</em>{v0}$, where $\rho_{v0}$ is a positive constant.

  • a) Use Gauss's Law to find the electric field intensity as a function of $r$.
  • b) Determine the uniform, cylindrical surface charge density required at a distance $r = b$ (b > a) to ensure that the electric field $E(r) = 0$ for r > b.

• Solution:

  • a) Applying Gauss's Law:
    • Gaussian surface: Cylinder with radius $r$ (through point P).
    • From symmetry: $E(P) = E(r) \hat{r}$.
      $\oint<em>S D \cdot dS = \int</em>V \rho dV$
      $\epsilon<em>0 \oint</em>S E \cdot dS = \int<em>V \rho dV$$\epsilon</em>0 E(r) 2 \pi r L = \int<em>0^r \int</em>0^L \int<em>0^{2 \pi} \rho</em>{v0} r dr d\phi dz$
      $\epsilon<em>0 E(r) 2 \pi r L = \rho</em>{v0} \pi r^2 L$
      $E(r) = \frac{\rho<em>{v0} r}{2 \epsilon</em>0} \hat{r} \quad (V/m)$
  • r > a:
    $\epsilon<em>0 E(r) 2 \pi r L = \int</em>0^a \int<em>0^L \int</em>0^{2 \pi} \rho<em>{v0} r dr d\phi dz$$E(r) = \frac{\rho</em>{v0} a^2}{2 \epsilon_0 r} \hat{r}$
  • b) For r > b: E(P) = 0
    $Q<em>v + Q</em>s = 0$
    $\int<em>V \rho dV + \int</em>S \rho<em>s dS = 0$$\rho</em>{vo} \pi a^2 L + \rho<em>s 2 \pi b L = 0$$\rho</em>s = - \frac{\rho_{vo} a^2 }{2 b}$

2012 Test 2 Q2

• Problem: A spherical charge distribution with radius $R = a$ is centered at the origin. The charge is uniformly distributed throughout the sphere with a volume charge density $\rho_v$ C/m³.

• a) Find the electric field intensity for the region R > a.

• Solution:

  • Applying Gauss's law. Gaussian surface: Sphere with radius R.
    $\oint<em>S D \cdot dS = Q</em>{incl} = \int_V \rho dV$
  • Region 1: For R > a
    $\oint D \cdot dS = \int \rho<em>v dV$$E(R) 4 \pi R^2 = \frac{\rho</em>v 4 \pi a^3}{3 \epsilon<em>0}$$E = \frac{\rho</em>v a^3}{3 \epsilon_0 R^2} \hat{R}$

• c) If the charge distribution in Figure Q2.1 contains a spherical air-filled cavity with radius R = b as shown in Figure Q2.2, determine the electric field intensity at the point c on the x-axis.

• Solution:
  $E = E<em>{sphere} + E</em>{cavity}$
  $E = \frac{\rho<em>v a^3}{3 \epsilon</em>0 R^2} - \frac{\rho<em>v b^3}{3 \epsilon</em>0 R^2}$
  $E(x) = \frac{\rho<em>v}{3 \epsilon</em>0} (\frac{ a^3}{c^2} - \frac{ b^3}{(b-c)^2})$