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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:

  1. Like charges repel, opposite charges attract
  2. Charge is quantized: Comes in multiples of elementary charge e=1.60×10−19 Ce = 1.60 \times 10^{-19} \text{ C}
  3. Charge is conserved: Total charge in isolated system is constant
  4. Charge is invariant: Doesn't depend on reference frame

Units of Charge

Coulomb (C): SI unit of electric charge

Elementary charge: e=1.60×10−19e = 1.60 \times 10^{-19} C

  • Charge of proton: +e+e
  • Charge of electron: −e-e

Any charge: q=neq = ne where nn is an integer

Charging Methods

1. Friction (Triboelectric Effect)

  • Rubbing transfers electrons
  • Example: Rubbing balloon on hair

2. Conduction (Contact)

  • Direct contact transfers charge
  • Charge distributes between objects

3. Induction

  • Charge separation without contact
  • Grounding removes one type of charge

Conductors vs. Insulators

Conductors

  • Allow charges to move freely
  • Examples: Metals, graphite, salt water
  • Excess charge distributes on surface

Insulators (Dielectrics)

  • Charges cannot move freely
  • Examples: Rubber, glass, plastic, wood
  • Charge stays where placed

Semiconductors

  • Properties between conductors and insulators
  • Examples: Silicon, germanium

Coulomb's Law

Coulomb's Law gives the electrostatic force between two point charges:

F=k∣q1q2∣r2F = k\frac{|q_1 q_2|}{r^2}

where:

  • FF = electrostatic force (N)
  • k=8.99×109k = 8.99 \times 10^9 N·m²/C² (Coulomb's constant)
  • q1,q2q_1, q_2 = charges (C)
  • rr = distance between charges (m)

Vector Form:

F⃗=kq1q2r2r^\vec{F} = k\frac{q_1 q_2}{r^2}\hat{r}

Direction:

  • Same signs → repulsive (away from each other)
  • Opposite signs → attractive (toward each other)

Coulomb's Constant

k=14πϵ0=8.99×109 N⋅m2/C2k = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 \text{ N·m}^2\text{/C}^2

where ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C²/(N·m²) is the permittivity of free space.

Often approximated as: k≈9.0×109k \approx 9.0 \times 10^9 N·m²/C²

Comparison with Gravity

Similarities:

  • Both are inverse square laws: F∝1/r2F \propto 1/r^2
  • Both are long-range forces
  • Both act along line connecting objects

Differences:

PropertyGravityElectricity
AlwaysAttractiveAttractive OR repulsive
ConstantG=6.67×10−11G = 6.67 \times 10^{-11}k=8.99×109k = 8.99 \times 10^9
Relative strengthVery weakMuch stronger
ShieldingNoYes (conductors)

Electric force is ~10³⁹ times stronger than gravity!

Superposition Principle

For multiple charges, total force = vector sum of individual forces:

F⃗total=F⃗1+F⃗2+F⃗3+...\vec{F}_{total} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ...

Each force calculated using Coulomb's Law independently.

Steps:

  1. Calculate force from each charge separately
  2. Determine direction of each force
  3. Resolve into components if needed
  4. Add vectors (components)

Problem-Solving Strategy

  1. Draw diagram showing all charges and distances
  2. Identify charge signs (+ or -)
  3. Apply Coulomb's Law for each pair
  4. Determine force directions:
    • Like charges → repel
    • Unlike charges → attract
  5. Use superposition for multiple charges
  6. Add as vectors (use components if needed)

Common Mistakes

❌ Forgetting to square the distance in denominator ❌ Using wrong sign convention (use magnitude, then add direction) ❌ Not converting units (must use meters, coulombs) ❌ Adding forces as scalars instead of vectors ❌ Confusing force on A from B with force on B from A (equal magnitude, opposite direction)

📚 Practice Problems

1Problem 1easy

❓ 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)

💡 Show Solution

Given:

  • q1=+3.0q_1 = +3.0 μC =3.0×10−6= 3.0 \times 10^{-6} C
  • q2=−2.0q_2 = -2.0 μC =−2.0×10−6= -2.0 \times 10^{-6} C
  • r=0.50r = 0.50 m
  • k=9.0×109k = 9.0 \times 10^9 N·m²/C²

Find: Force FF

Solution:

Apply Coulomb's Law: F=k∣q1q2∣r2F = k\frac{|q_1 q_2|}{r^2} F=(9.0×109)∣(3.0×10−6)(−2.0×10−6)∣(0.50)2F = (9.0 \times 10^9)\frac{|(3.0 \times 10^{-6})(-2.0 \times 10^{-6})|}{(0.50)^2} F=(9.0×109)6.0×10−120.25F = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.25} F=(9.0×109)(2.4×10−11)F = (9.0 \times 10^9)(2.4 \times 10^{-11}) F=0.216 N≈0.22 NF = 0.216 \text{ N} \approx 0.22 \text{ N}

Direction: Opposite signs → attractive force

Answer: 0.22 N, attractive

2Problem 2easy

❓ 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)

💡 Show Solution

Given:

  • q1=+3.0q_1 = +3.0 μC =3.0×10−6= 3.0 \times 10^{-6} C
  • q2=−2.0q_2 = -2.0 μC =−2.0×10−6= -2.0 \times 10^{-6} C
  • r=0.50r = 0.50 m
  • k=9.0×109k = 9.0 \times 10^9 N·m²/C²

Find: Force FF

Solution:

Apply Coulomb's Law: F=k∣q1q2∣r2F = k\frac{|q_1 q_2|}{r^2} F=(9.0×109)∣(3.0×10−6)(−2.0×10−6)∣(0.50)2F = (9.0 \times 10^9)\frac{|(3.0 \times 10^{-6})(-2.0 \times 10^{-6})|}{(0.50)^2} F=(9.0×109)6.0×10−120.25F = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.25} F=(9.0×109)(2.4×10−11)F = (9.0 \times 10^9)(2.4 \times 10^{-11}) F=0.216 N≈0.22 NF = 0.216 \text{ N} \approx 0.22 \text{ N}

Direction: Opposite signs → attractive force

Answer: 0.22 N, attractive

3Problem 3medium

❓ 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?

💡 Show Solution

Given:

  • q1=+2.0×10−6q_1 = +2.0 \times 10^{-6} C at x = 0
  • q2=−3.0×10−6q_2 = -3.0 \times 10^{-6} C at x = 0.30 m
  • q3=+1.0×10−6q_3 = +1.0 \times 10^{-6} C at x = 0.60 m

Find: Net force on q2q_2

Solution:

Force from q₁ on q₂:

Distance: r12=0.30r_{12} = 0.30 m F12=k∣q1q2∣r122=(9.0×109)(2.0×10−6)(3.0×10−6)(0.30)2F_{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} F12=(9.0×109)6.0×10−120.09=0.60 NF_{12} = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.09} = 0.60 \text{ N}

Direction: Opposite signs → attractive → toward q₁ (left, negative) F⃗12=−0.60 N\vec{F}_{12} = -0.60 \text{ N}

Force from q₃ on q₂:

Distance: r23=0.60−0.30=0.30r_{23} = 0.60 - 0.30 = 0.30 m F23=k∣q2q3∣r232=(9.0×109)(3.0×10−6)(1.0×10−6)(0.30)2F_{23} = k\frac{|q_2 q_3|}{r_{23}^2} = (9.0 \times 10^9)\frac{(3.0 \times 10^{-6})(1.0 \times 10^{-6})}{(0.30)^2} F23=(9.0×109)3.0×10−120.09=0.30 NF_{23} = (9.0 \times 10^9)\frac{3.0 \times 10^{-12}}{0.09} = 0.30 \text{ N}

Direction: Opposite signs → attractive → toward q₃ (right, positive) F⃗23=+0.30 N\vec{F}_{23} = +0.30 \text{ N}

Net force (superposition): Fnet=F12+F23=−0.60+0.30=−0.30 NF_{net} = F_{12} + F_{23} = -0.60 + 0.30 = -0.30 \text{ N}

Answer: 0.30 N to the left (toward q₁)

4Problem 4medium

❓ 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?

💡 Show Solution

Given:

  • q1=+2.0×10−6q_1 = +2.0 \times 10^{-6} C at x = 0
  • q2=−3.0×10−6q_2 = -3.0 \times 10^{-6} C at x = 0.30 m
  • q3=+1.0×10−6q_3 = +1.0 \times 10^{-6} C at x = 0.60 m

Find: Net force on q2q_2

Solution:

Force from q₁ on q₂:

Distance: r12=0.30r_{12} = 0.30 m F12=k∣q1q2∣r122=(9.0×109)(2.0×10−6)(3.0×10−6)(0.30)2F_{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} F12=(9.0×109)6.0×10−120.09=0.60 NF_{12} = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.09} = 0.60 \text{ N}

Direction: Opposite signs → attractive → toward q₁ (left, negative) F⃗12=−0.60 N\vec{F}_{12} = -0.60 \text{ N}

Force from q₃ on q₂:

Distance: r23=0.60−0.30=0.30r_{23} = 0.60 - 0.30 = 0.30 m F23=k∣q2q3∣r232=(9.0×109)(3.0×10−6)(1.0×10−6)(0.30)2F_{23} = k\frac{|q_2 q_3|}{r_{23}^2} = (9.0 \times 10^9)\frac{(3.0 \times 10^{-6})(1.0 \times 10^{-6})}{(0.30)^2} F23=(9.0×109)3.0×10−120.09=0.30 NF_{23} = (9.0 \times 10^9)\frac{3.0 \times 10^{-12}}{0.09} = 0.30 \text{ N}

Direction: Opposite signs → attractive → toward q₃ (right, positive) F⃗23=+0.30 N\vec{F}_{23} = +0.30 \text{ N}

Net force (superposition): Fnet=F12+F23=−0.60+0.30=−0.30 NF_{net} = F_{12} + F_{23} = -0.60 + 0.30 = -0.30 \text{ N}

Answer: 0.30 N to the left (toward q₁)

5Problem 5medium

❓ 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.

💡 Show Solution

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)

6Problem 6hard

❓ 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?

💡 Show Solution

Given:

  • Four charges: Q=5.0×10−6Q = 5.0 \times 10^{-6} C each
  • Square side: a=0.20a = 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 = a2a\sqrt{2} (diagonal)

Force from adjacent charges (two): Fadj=kQ2a2=(9.0×109)(5.0×10−6)2(0.20)2F_{adj} = k\frac{Q^2}{a^2} = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})^2}{(0.20)^2} Fadj=(9.0×109)2.5×10−110.04=5.625 NF_{adj} = (9.0 \times 10^9)\frac{2.5 \times 10^{-11}}{0.04} = 5.625 \text{ N}

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

Force from diagonal charge: Fdiag=kQ2(a2)2=kQ22a2=Fadj2=2.813 NF_{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: F⃗1=(5.625,0)\vec{F}_1 = (5.625, 0) N From top: F⃗2=(0,5.625)\vec{F}_2 = (0, 5.625) 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: Fx=5.625+0+1.99=7.62 NF_x = 5.625 + 0 + 1.99 = 7.62 \text{ N} Fy=0+5.625+1.99=7.62 NF_y = 0 + 5.625 + 1.99 = 7.62 \text{ N}

Fnet=Fx2+Fy2=7.622+7.622=10.8 NF_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{7.62^2 + 7.62^2} = 10.8 \text{ N}

θ=tan⁡−1FyFx=45°\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!

7Problem 7hard

❓ 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.

💡 Show Solution

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

8Problem 8hard

❓ 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?

💡 Show Solution

Given:

  • Four charges: Q=5.0×10−6Q = 5.0 \times 10^{-6} C each
  • Square side: a=0.20a = 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 = a2a\sqrt{2} (diagonal)

Force from adjacent charges (two): Fadj=kQ2a2=(9.0×109)(5.0×10−6)2(0.20)2F_{adj} = k\frac{Q^2}{a^2} = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})^2}{(0.20)^2} Fadj=(9.0×109)2.5×10−110.04=5.625 NF_{adj} = (9.0 \times 10^9)\frac{2.5 \times 10^{-11}}{0.04} = 5.625 \text{ N}

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

Force from diagonal charge: Fdiag=kQ2(a2)2=kQ22a2=Fadj2=2.813 NF_{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: F⃗1=(5.625,0)\vec{F}_1 = (5.625, 0) N From top: F⃗2=(0,5.625)\vec{F}_2 = (0, 5.625) 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: Fx=5.625+0+1.99=7.62 NF_x = 5.625 + 0 + 1.99 = 7.62 \text{ N} Fy=0+5.625+1.99=7.62 NF_y = 0 + 5.625 + 1.99 = 7.62 \text{ N}

Fnet=Fx2+Fy2=7.622+7.622=10.8 NF_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{7.62^2 + 7.62^2} = 10.8 \text{ N}

θ=tan⁡−1FyFx=45°\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!

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