Electrostatics and Image Theory

Coulomb's Law

Electric field intensity E is defined as: E = \frac{q}{4 \pi \epsilon_0 R^2} \hat{R}
- Units: Volts per meter (V/m)
Electric potential V(R) is constant on a Gaussian surface.
Gaussian surface is a sphere.

The image method simplifies calculations for E and V due to charges near conducting planes.
Steps:
1. For each charge Q, add an image charge -Q.
2. Remove the conducting plane.
3. Calculate the field due to all charges.

Problem: Determine E at an arbitrary point P = (x, y, z) in the region z > 0 due to a charge Q in free space at a distance d above a grounded conducting plate residing in the z = 0 plane.
Solution:
- Charge Q is at (0, 0, d), and its image -Q is at (0, 0, -d).
- Using the formula:
  E = \frac{1}{4 \pi \epsilon0} \left[ \frac{Q R1}{R1^3} - \frac{Q R2}{R_2^3} \right]
- The electric field at point P = (x, y, z) due to the two charges is:
  E = \frac{Q}{4 \pi \epsilon_0} \left[ \frac{\hat{x}x + \hat{y}y + \hat{z}(z - d)}{[x^2 + y^2 + (z - d)^2]^{3/2}} - \frac{\hat{x}x + \hat{y}y + \hat{z}(z + d)}{[x^2 + y^2 + (z + d)^2]^{3/2}} \right]
  for z \geq 0.

A spherical shell with radius a is at a distance d above a perfect conducting ground plane and contains a surface charge density \rho_s C/m². Determine the electric field intensity at any point on the contour described by x = 0, z = d.
When R > a:
E = E{R1} = \frac{\rhos a^2}{ \epsilon0 R^2} \int \vec{E} \cdot d\vec{S} = \int \frac{\rhos}{\epsilon0} dS E = \frac{\rhos}{ \epsilon0} \frac{a^2}{R^2} ER = \frac{\rhos a^2}{\epsilon0 R^2}
\int \int \vec{E} \cdot d\vec{S} = \frac{\rhos \pi a^4}{ \epsilon0}
When R < a:
\int \int{S} \vec{ER} \cdot d\vec{S} = 0
With image charge:
E = \int \int - \frac{\rhos}{ \epsilon0} dS

Case 1: R > a
E1 = \frac{\rhos a^2}{\epsilon0 R^2} \hat{R} E2 = -\frac{\rhos a^2}{\epsilon0 R^2} \hat{R}
ET = E1 + E_2
Case 2: R > a
E = \frac{\rhos}{ \epsilon0} \frac{a^2}{R^2} \hat{y}

Determine the electric field intensity E at any point on the contour described by x = 0, z = d.

Problem: A spherical shell with radius a is at a distance d above a perfect conducting ground plane with surface charge density \rho_0 C/m². Determine:
- a) the induced charge density on the ground plane
- b) the total induced charge on the ground plane

Use Image Theory:
- E2 = D2 = 0
- \hat{n2} \cdot (D1 - D2) = \rhos
- E2 - E1 = \frac{\rhos}{\epsilon0} \hat{n_2}

E = E1 + E2

Electric field due to original charge distribution:
E_1
Electric field due to image charge distribution:
E_2
For point P(r, \phi, 0), which is in the z = 0 plane:
E = \frac{\rho}{\epsilon0} \frac{RP}{R^2} \hat{R}

E1 (P) = \frac{\rho0}{ \epsilon} \frac{a^2}{R^2} \hat{R}

The electric field for the original charge Q at distance d and image charge -Q with E1 and E2 is given by:
E = \frac{\rho}{ \epsilon0} [cos(\alpha) \hat{r} - sin(\alpha) \hat{z}] E = \frac{\rho}{ \epsilon0} [sin(\alpha) \hat{r} + cos(\alpha) \hat{z}]
Since D2 = 0, \rhos = -\hat{n2} \cdot D1
The induced charge density is given by:
\rhos = \frac{\rho}{ \epsilon0} \frac{a^2 d}{(r^2 + d^2 + a^2)^{3/2}}

Integrate induced charge density over the surface to find the total induced charge.
\int \int \rhos dS = Q{induced}
Q{induced} = \int{0}^{2 \pi} \int_{0}^{\infty} \frac{\rho}{[r^2 + (d-a)^2]^{3/2}} r dr d\phi
After integration:
Q_{induced} = -4 \pi a \rho

A similar setup with parameters d, a, \rho_s, and \sigma = \infty.
C = \frac{Q}{V_{ab}}
- Relationship between electric field and potential:
  V_{ab} = - \int E \cdot dl
The total electric field ET is given by ET = E(z) = \frac{\rhos}{ \epsilon0} \left[ \frac{d-z}{[a^2 + (d-z)^2]^{3/2}} - \frac{d+z}{[a^2 + (d+z)^2]^{3/2}} \right] \hat{z}
V = \frac{Q}{C} = \frac{4 \pi \rho [\sqrt{a^2 + (d-a)^2} - \sqrt{a^2 + (d+a)^2 }]}{4 \pi \epsilon_0 \epsilon}