What is Electric Charge and Coulomb's Law?

Properties of electric charge, conservation, and electrostatic forces

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⚡ Electric Charge and Coulomb's Law

Electric Charge

Electric charge is a fundamental property of matter that causes electromagnetic interactions.

Two Types:

Positive (+): Protons
Negative (-): Electrons

Key Properties:

Like charges repel, opposite charges attract
Charge is quantized: Comes in multiples of elementary charge $e = 1.60 \times 10^{-19} \text{ C}$

❓ Question:

Two point charges, q₁ = +3.0 μC and q₂ = -2.0 μC, are separated by 0.50 m. What is the magnitude of the electrostatic force between them? (k = 9.0 × 10⁹ N·m²/C²)

Given:

$q_1 = +3.0$ μC C

Property	Gravity	Electricity
Always	Attractive	Attractive OR repulsive
Constant	$G = 6.67 \times 10^{-11}$	$k = 8.99 \times 10^9$
Relative strength	Very weak	Much stronger
Shielding	No	Yes (conductors)

❓ Question:

Two point charges, q₁ = +3.0 μC and q₂ = -2.0 μC, are separated by 0.50 m. What is the magnitude of the electrostatic force between them? (k = 9.0 × 10⁹ N·m²/C²)

Given:

$q_1 = +3.0$ μC $= 3.0 \times 10^{-6}$ C
$q_2 = -2.0$ μC $= -2.0 \times 10^{-6}$ C
$r = 0.50$ m
$k = 9.0 \times 10^9$ N·m²/C²

Find: Force $F$

Solution:

Apply Coulomb's Law: $F = k\frac{|q_1 q_2|}{r^2}$

Direction: Opposite signs → attractive force

Answer: 0.22 N, attractive

❓ Question:

Three charges are arranged in a line: q₁ = +2.0 μC at x = 0, q₂ = -3.0 μC at x = 0.30 m, and q₃ = +1.0 μC at x = 0.60 m. What is the net force on q₂?

Given:

$q_1 = +2.0 \times 10^{-6}$ C at x = 0
$q_2 = -3.0 \times 10^{-6}$ C at x = 0.30 m
$q_3 = +1.0 \times 10^{-6}$ C at x = 0.60 m

Find: Net force on $q_2$

Solution:

Force from q₁ on q₂:

Distance: $r_{12} = 0.30$ m $F_{12} = k\frac{|q_1 q_2|}{r_{12}^2} = (9.0 \times 10^9)\frac{(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.30)^2}$

Direction: Opposite signs → attractive → toward q₁ (left, negative) $\vec{F}_{12} = -0.60 \text{ N}$

Force from q₃ on q₂:

Distance: $r_{23} = 0.60 - 0.30 = 0.30$ m $F_{23} = k \frac{∣ q_{2} q_{3} ∣}{r_{23}^{2}} = (9.0 \times 10^{9}) \frac{(3.0 \times}{}$

Direction: Opposite signs → attractive → toward q₃ (right, positive) $\vec{F}_{23} = +0.30 \text{ N}$

Net force (superposition): $F_{net} = F_{12} + F_{23} = -0.60 + 0.30 = -0.30 \text{ N}$

Answer: 0.30 N to the left (toward q₁)

❓ Question:

Three charges are arranged in a line: q₁ = +2.0 μC at x = 0, q₂ = -3.0 μC at x = 0.30 m, and q₃ = +1.0 μC at x = 0.60 m. What is the net force on q₂?

Given:

$q_1 = +2.0 \times 10^{-6}$ C at x = 0
$q_2 = -3.0 \times 10^{-6}$ C at x = 0.30 m
$q_3 = +1.0 \times 10^{-6}$ C at x = 0.60 m

Find: Net force on $q_2$

Solution:

Force from q₁ on q₂:

Distance: $r_{12} = 0.30$ m $F_{12} = k\frac{|q_1 q_2|}{r_{12}^2} = (9.0 \times 10^9)\frac{(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.30)^2}$

Direction: Opposite signs → attractive → toward q₁ (left, negative) $\vec{F}_{12} = -0.60 \text{ N}$

Force from q₃ on q₂:

Distance: $r_{23} = 0.60 - 0.30 = 0.30$ m $F_{23} = k \frac{∣ q_{2} q_{3} ∣}{r_{23}^{2}} = (9.0 \times 10^{9}) \frac{(3.0 \times}{}$

Direction: Opposite signs → attractive → toward q₃ (right, positive) $\vec{F}_{23} = +0.30 \text{ N}$

Net force (superposition): $F_{net} = F_{12} + F_{23} = -0.60 + 0.30 = -0.30 \text{ N}$

Answer: 0.30 N to the left (toward q₁)

❓ Question:

Two point charges, q₁ = +3.0 μC and q₂ = -5.0 μC, are separated by 0.20 m. (a) What is the magnitude of the electric force between them? (b) Is the force attractive or repulsive? Use k = 9.0 × 10⁹ N·m²/C².

Solution:

Given: q₁ = +3.0 × 10⁻⁶ C, q₂ = -5.0 × 10⁻⁶ C, r = 0.20 m, k = 9.0 × 10⁹ N·m²/C²

(a) Force magnitude (Coulomb's Law): F = k|q₁q₂|/r² F = (9.0 × 10⁹)|3.0 × 10⁻⁶ × (-5.0 × 10⁻⁶)|/(0.20)² F = (9.0 × 10⁹)(15 × 10⁻¹²)/0.04 F = (135 × 10⁻³)/0.04 F = 3.4 N

(b) Attractive or repulsive? q₁ is positive, q₂ is negative (opposite signs) Attractive force (opposite charges attract)

❓ Question:

Four equal charges Q = +5.0 μC are placed at the corners of a square with side length 0.20 m. What is the magnitude and direction of the net force on one of the charges?

Given:

Four charges: $Q = 5.0 \times 10^{-6}$ C each
Square side: $a = 0.20$ m

Find: Net force on one corner charge

Solution:

Consider force on charge at origin (0,0). Other charges at:

(a, 0): distance = a
(0, a): distance = a
(a, a): distance = $a\sqrt{2}$ (diagonal)

Force from adjacent charges (two): $F_{adj} = k\frac{Q^2}{a^2} = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})^2}{(0.20)^2}$

Each points along edge (90° angle between them).

Force from diagonal charge: $F_{diag} = k\frac{Q^2}{(a\sqrt{2})^2} = k\frac{Q^2}{2a^2} = \frac{F_{adj}}{2} = 2.813 \text{ N}$

Points along diagonal toward corner.

Components:

From right: $\vec{F}_1 = (5.625, 0)$ N From top: N From diagonal: At 45°, points toward (along diagonal from origin) $$\vec{F}_3 = (2.813\cos 45°, 2.813\sin 45°) = (1.99, 1.99)$ N

Total: $F_x = 5.625 + 0 + 1.99 = 7.62 \text{ N}$ $F_{y} = 0 + 5.625 + 1.99 = 7.62 N$

$F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{7.62^2 + 7.62^2} = 10.8 \text{ N}$

$\theta = \tan^{-1}\frac{F_y}{F_x} = 45°$

Answer: 10.8 N at 45° from the two adjacent sides (along diagonal outward)

By symmetry, this makes sense - force points directly away from center!

❓ Question:

Three charges are arranged in a line: q₁ = +2.0 μC at x = 0, q₂ = -4.0 μC at x = 0.30 m, and q₃ = +1.0 μC at x = 0.50 m. Find the net force on q₂. Use k = 9.0 × 10⁹ N·m²/C².

Solution:

Force on q₂ from q₁: r₁₂ = 0.30 m F₁₂ = kq₁q₂/r₁₂² = (9.0 × 10⁹)(2.0 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.30)² F₁₂ = (72 × 10⁻³)/0.09 = 0.800 N

Direction: q₁ (+) attracts q₂ (-) → 0.800 N to the left (negative direction)

Force on q₂ from q₃: r₂₃ = 0.50 - 0.30 = 0.20 m F₂₃ = kq₂q₃/r₂₃² = (9.0 × 10⁹)(4.0 × 10⁻⁶)(1.0 × 10⁻⁶)/(0.20)² F₂₃ = (36 × 10⁻³)/0.04 = 0.900 N

Direction: q₃ (+) attracts q₂ (-) → 0.900 N to the right (positive direction)

Net force: F_net = -0.800 + 0.900 = +0.100 N or 0.10 N to the right

❓ Question:

Four equal charges Q = +5.0 μC are placed at the corners of a square with side length 0.20 m. What is the magnitude and direction of the net force on one of the charges?

Given:

Four charges: $Q = 5.0 \times 10^{-6}$ C each
Square side: $a = 0.20$ m

Find: Net force on one corner charge

Solution:

Consider force on charge at origin (0,0). Other charges at:

(a, 0): distance = a
(0, a): distance = a
(a, a): distance = $a\sqrt{2}$ (diagonal)

Force from adjacent charges (two): $F_{adj} = k\frac{Q^2}{a^2} = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})^2}{(0.20)^2}$

Each points along edge (90° angle between them).

Force from diagonal charge: $F_{diag} = k\frac{Q^2}{(a\sqrt{2})^2} = k\frac{Q^2}{2a^2} = \frac{F_{adj}}{2} = 2.813 \text{ N}$

Points along diagonal toward corner.

Components:

From right: $\vec{F}_1 = (5.625, 0)$ N From top: N From diagonal: At 45°, points toward (along diagonal from origin) $$\vec{F}_3 = (2.813\cos 45°, 2.813\sin 45°) = (1.99, 1.99)$ N

Total: $F_x = 5.625 + 0 + 1.99 = 7.62 \text{ N}$ $F_{y} = 0 + 5.625 + 1.99 = 7.62 N$

$F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{7.62^2 + 7.62^2} = 10.8 \text{ N}$

$\theta = \tan^{-1}\frac{F_y}{F_x} = 45°$

Answer: 10.8 N at 45° from the two adjacent sides (along diagonal outward)

By symmetry, this makes sense - force points directly away from center!

Electric Charge and Coulomb's Law

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⚡ Electric Charge and Coulomb's Law

Electric Charge

Two Types:

Key Properties:

📚 Practice Problems

1Problem 1easy

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📌 Related Topics in Electrostatics

❓ Frequently Asked Questions

Units of Charge

Charging Methods

1. Friction (Triboelectric Effect)

2. Conduction (Contact)

3. Induction

Conductors vs. Insulators

Conductors

Insulators (Dielectrics)

Semiconductors

Coulomb's Law

Vector Form:

Coulomb's Constant

Comparison with Gravity

Superposition Principle

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