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Conductors in Electric Fields

Conductors are materials that allow free movement of electrons. When placed in electric fields, conductors exhibit unique behaviors that are fundamental to understanding electrical systems and electrostatic shielding.

What is a Conductor?

Definition

A conductor is a material that contains free electrons that can move easily through the material. Metals like copper, aluminum, and silver are excellent conductors.

In conductors, some electrons are not bound to individual atoms but can move freely throughout the material. This free electron movement is what makes conductors able to carry electric current and respond to electric fields.

Electrons moving in a conductor
Free electrons in a conductor can move throughout the material.

Electrostatic Equilibrium

Key Principle

In electrostatic equilibrium, the electric field inside a conductor is zero:

$$\vec{E}_{\text{inside}} = 0$$

When a conductor is placed in an external electric field, the free electrons redistribute themselves until electrostatic equilibrium is reached. At this point:

Properties of Conductors in Electric Fields

Fundamental Properties

Charge Distribution on Conductors

Surface Charge Density

The charge on a conductor's surface is distributed according to the surface curvature:

$$\sigma = \frac{Q}{A}$$

Where σ is surface charge density, Q is total charge, and A is surface area.

Curvature Effect

Charge density is higher on sharp points and lower on flat surfaces:

Charge Distribution on Conductor
Charge density varies with surface curvature.

Electric Field at Conductor Surface

Surface Field Formula

The electric field just outside a conductor surface is:

$$E = \frac{\sigma}{\epsilon_0}$$

Where σ is the surface charge density and ε₀ is the permittivity of free space.

This relationship shows that:

Worked Examples

Example 1: Electric Field Inside a Conductor

Problem: A conducting sphere of radius 5.0 cm has a total charge of +2.0 μC. What is the electric field inside the sphere?

Solution Steps:

  1. Given: Q = +2.0 × 10⁻⁶ C, r = 5.0 × 10⁻² m
  2. Property: Electric field inside conductor is zero
  3. Answer: E = 0 N/C

Answer: The electric field inside the conductor is zero (E = 0 N/C).

Example 2: Surface Charge Density

Problem: A conducting sphere of radius 3.0 cm has a total charge of -1.5 μC. What is the surface charge density?

Solution Steps:

  1. Given: Q = -1.5 × 10⁻⁶ C, r = 3.0 × 10⁻² m
  2. Surface area: A = 4πr² = 4π(3.0 × 10⁻²)² = 1.13 × 10⁻² m²
  3. Surface charge density: σ = Q/A = (-1.5 × 10⁻⁶)/(1.13 × 10⁻²)
  4. Calculate: σ = -1.33 × 10⁻⁴ C/m²

Answer: The surface charge density is -1.33 × 10⁻⁴ C/m² (negative because the charge is negative).

Example 3: Electric Field at Conductor Surface

Problem: A conducting plate has a surface charge density of +3.0 × 10⁻⁶ C/m². What is the electric field just outside the plate?

Solution Steps:

  1. Given: σ = +3.0 × 10⁻⁶ C/m², ε₀ = 8.85 × 10⁻¹² C²/N⋅m²
  2. Formula: E = σ/ε₀
  3. Substitute: E = (3.0 × 10⁻⁶)/(8.85 × 10⁻¹²)
  4. Calculate: E = 3.39 × 10⁵ N/C

Answer: The electric field just outside the plate is 3.39 × 10⁵ N/C, directed away from the plate (since the charge is positive).

Conductors in External Electric Fields

Induction Process

When a conductor is placed in an external electric field:

  1. Initial state: External field penetrates the conductor
  2. Electron movement: Free electrons move in response to the field
  3. Charge separation: Positive and negative charges separate
  4. Induced field: Induced charges create their own electric field
  5. Equilibrium: Induced field cancels external field inside conductor

Faraday Cage Effect

A hollow conductor completely shields its interior from external electric fields:

Conductor in External Field
A conductor in an external electric field reaches electrostatic equilibrium.

Applications of Conductors

Common Mistakes to Avoid

⚠️ Common Errors

Practice Problems

Practice Problem 1

Problem: A conducting sphere of radius 4.0 cm has a surface charge density of +2.0 × 10⁻⁶ C/m². What is the electric field just outside the sphere?

Click for solution

Solution:

  1. Given: σ = +2.0 × 10⁻⁶ C/m², ε₀ = 8.85 × 10⁻¹² C²/N⋅m²
  2. Formula: E = σ/ε₀
  3. Substitute: E = (2.0 × 10⁻⁶)/(8.85 × 10⁻¹²)
  4. Calculate: E = 2.26 × 10⁵ N/C

Answer: 2.26 × 10⁵ N/C, directed away from the sphere

Practice Problem 2

Problem: A conducting cube has a total charge of +6.0 μC. If the cube has side length 2.0 cm, what is the surface charge density?

Click for solution

Solution:

  1. Given: Q = +6.0 × 10⁻⁶ C, side length = 2.0 × 10⁻² m
  2. Surface area: A = 6 × (2.0 × 10⁻²)² = 2.4 × 10⁻³ m²
  3. Surface charge density: σ = Q/A = (6.0 × 10⁻⁶)/(2.4 × 10⁻³)
  4. Calculate: σ = 2.5 × 10⁻³ C/m²

Answer: 2.5 × 10⁻³ C/m²

Charge Sharing Between Conducting Spheres of Different Sizes

Key Principle

When two conducting spheres of different radii are brought into contact and then separated, the final charge on each sphere is not simply the average of their initial charges. Instead, charge distributes so that both spheres reach the same electric potential (voltage), because conductors in equilibrium are equipotentials.

For isolated spheres of radii R₁ and R₂ with initial charges q₁ and q₂:

Potential of a sphere: \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{R} \)

After contact and separation, set \( V_1 = V_2 \):

\( \frac{q_1'}{R_1} = \frac{q_2'}{R_2} \)

and \( q_1' + q_2' = q_1 + q_2 \) (conservation of charge)

Solving gives:

\( q_1' = (q_1 + q_2) \frac{R_1}{R_1 + R_2} \)

\( q_2' = (q_1 + q_2) \frac{R_2}{R_1 + R_2} \)

Diagram: Two spheres of different radii sharing charge
After contact, charge distributes so both spheres have the same potential.

Example: Charge Sharing for Different Sized Spheres

Problem: Sphere A (radius 2 cm) has +6 μC, Sphere B (radius 6 cm) has -2 μC. They are brought into contact and then separated. What is the final charge on each?

Solution Steps:

  1. Total charge: \( q_{\text{total}} = +6 \text{ μC} + (-2 \text{ μC}) = +4 \text{ μC} \)
  2. Radii: \( R_1 = 2 \text{ cm}, R_2 = 6 \text{ cm} \)
  3. Formula: \( q_1' = q_{\text{total}} \frac{R_1}{R_1 + R_2} \), \( q_2' = q_{\text{total}} \frac{R_2}{R_1 + R_2} \)
  4. Calculate: \( q_1' = 4 \times \frac{2}{2+6} = 1 \text{ μC} \), \( q_2' = 4 \times \frac{6}{2+6} = 3 \text{ μC} \)

Answer: After contact, Sphere A has +1 μC, Sphere B has +3 μC.

Key Concepts Summary

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