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 \rhol along the z-axis, the electric field at a point P(r, \phi, z) is: E(P) = \int \frac{dq}{4 \pi \epsilon0 R'^2} \hat{R} = \int \frac{\rhol dl'}{4 \pi \epsilon0 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{\rhol}{2 \pi \epsilon0 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:
  \ointS D \cdot ds = Q{encl}
  where:

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

• Divergence Theorem:
  \intV (\nabla \cdot D) dV = \ointS 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 (\epsilon0): \ointS D \cdot dS = Q{incl} = \intV \rho dV
  \ointS \epsilon0 E \cdot dS = Q_{incl}

• If there is no enclosed charge (Q{incl} = 0), then: \ointS D \cdot dS = 0

Applying Gauss's Law

• Electric Field due to a Single Charge:

  • Gaussian surface: Sphere of radius R.
  • \ointS E \cdot dS = E \ointS dS = E(4 \pi R^2) = \frac{Q}{\epsilon_0}

• Electric Field due to an Infinite Line Charge Density:

  • Gaussian surface: Cylinder.
  • E = \frac{\rhol}{2 \pi \epsilon0 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{S{top}} D \cdot dS = \int{S{bottom}} D \cdot dS = 0
  • The flux through the cylindrical surface is:
    \ointS \vec{D} \cdot d\vec{S} = Dr \oint dS = D_r (2 \pi r h)
  • Total charge enclosed:
    Q = \rho_l h
  • Applying Gauss's Law:
    Dr (2 \pi r h) = \rhol h
    D = \frac{\rhol}{2 \pi r} \hat{r} E = \frac{\rhol}{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 \rhov = \rho{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}.
      \ointS D \cdot dS = \intV \rho dV
      \epsilon0 \ointS E \cdot dS = \intV \rho dV \epsilon0 E(r) 2 \pi r L = \int0^r \int0^L \int0^{2 \pi} \rho{v0} r dr d\phi dz
      \epsilon0 E(r) 2 \pi r L = \rho{v0} \pi r^2 L
      E(r) = \frac{\rho{v0} r}{2 \epsilon0} \hat{r} \quad (V/m)
  • r > a:
    \epsilon0 E(r) 2 \pi r L = \int0^a \int0^L \int0^{2 \pi} \rho{v0} r dr d\phi dz E(r) = \frac{\rho{v0} a^2}{2 \epsilon_0 r} \hat{r}
  • b) For r > b: E(P) = 0
    Qv + Qs = 0
    \intV \rho dV + \intS \rhos dS = 0 \rho{vo} \pi a^2 L + \rhos 2 \pi b L = 0 \rhos = - \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.
    \ointS D \cdot dS = Q{incl} = \int_V \rho dV
  • Region 1: For R > a
    \oint D \cdot dS = \int \rhov dV E(R) 4 \pi R^2 = \frac{\rhov 4 \pi a^3}{3 \epsilon0} E = \frac{\rhov 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{sphere} + E{cavity}
  E = \frac{\rhov a^3}{3 \epsilon0 R^2} - \frac{\rhov b^3}{3 \epsilon0 R^2}
  E(x) = \frac{\rhov}{3 \epsilon0} (\frac{ a^3}{c^2} - \frac{ b^3}{(b-c)^2})