Principles of Electrostatics

An electric field is a region of space where a charged particle experiences a non-contact force. Coulomb's Law provides the mathematical framework for calculating the magnitude of this force between two stationary point charges.

F = (1 / 4πε₀) × (q₁q₂ / r²)

1. Coulomb's Law

The force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation distance. This is remarkably similar to Newton's Law of Gravitation, but the electric force can be either attractive or repulsive.

2. Electric Field Strength (E)

Electric field strength at a point is defined as the force per unit positive charge acting on a stationary point charge at that point. It is a vector quantity, pointing in the direction that a positive test charge would move.

E = F / q = Q / (4πε₀r²)

3. Electric Potential (V)

Unlike field strength, electric potential is a scalar. It is defined as the work done per unit positive charge in bringing a small test charge from infinity to the point in question.

V = Q / (4πε₀r)
Exam Tip: For any point between two identical positive charges, the electric field strength will be zero at the midpoint because the vectors cancel out. However, the electric potential will NOT be zero, as potentials from multiple charges simply add up algebraically.

Deep Dive: Worked Examples

✅ Example 1: Force between Protons

Calculate the electrostatic force between two protons in a nucleus, separated by 2.0 × 10⁻¹⁵ m.

Step 1: Identify values q₁ = q₂ = +1.6 × 10⁻¹⁹ C, r = 2.0 × 10⁻¹⁵ m, k ≈ 9.0 × 10⁹
Step 2: Apply F = kq₁q₂ / r² F = (9.0 × 10⁹) × (1.6e-19)² / (2.0e-15)²
Step 3: Solve F = (9.0e9 × 2.56e-38) / 4.0e-30 = 57.6 N (Repulsive)

✅ Example 2: Field Strength of a Van de Graaff

A Van de Graaff sphere has a charge of 3.0 μC. Find the field strength 20 cm from its center.

Step 1: Convert units Q = 3.0 × 10⁻⁶ C, r = 0.20 m
Step 2: Apply E = kQ / r² E = (9.0 × 10⁹) × (3.0e-6) / (0.20)²
Step 3: Solve E = 27000 / 0.04 = 6.75 × 10⁵ V/m

✅ Example 3: Work Done (Energy)

How much work is required to move an electron (q = -1.6e-19) through a potential difference of 5000 V?

Step 1: Formula W = qΔV
Step 2: Calculate W = (1.6 × 10⁻¹⁹) × 5000
Step 3: Solve W = 8.0 × 10⁻¹⁶ J

✅ Example 4: Alpha Particle Scattering

An Alpha particle (+2e) is 1.0 × 10⁻¹² m from a Gold nucleus (+79e). Find the potential energy of the system.

Step 1: Formula for EPE EPE = kq₁q₂ / r
Step 2: Plug in values EPE = (9e9 × 2 × 1.6e-19 × 79 × 1.6e-19) / 1e-12
Step 3: Solve EPE ≈ 3.64 × 10⁻¹⁴ J

✅ Example 5: Capacitance & Field

Two parallel plates 5mm apart have a P.D. of 200V. Calculate the uniform field strength between them.

Step 1: Formula for uniform field E = V / d
Step 2: Convert units d = 0.005 m
Step 3: Solve E = 200 / 0.005 = 40,000 V/m