Electric Charge and Coulomb's Law

Properties of electric charge, conservation, and electrostatic forces

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โšก 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=1.60ร—10โˆ’19ย 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=1.60ร—10โˆ’19e = 1.60 \times 10^{-19} C

  • Charge of proton: +e+e
  • Charge of electron: โˆ’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=kโˆฃq1q2โˆฃr2F = k\frac{|q_1 q_2|}{r^2}

where:

  • FF = electrostatic force (N)
  • k=8.99ร—109k = 8.99 \times 10^9 Nยทmยฒ/Cยฒ (Coulomb's constant)
  • q1,q2q_1, q_2 = charges (C)
  • rr = distance between charges (m)

Vector Form:

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

Direction:

  • Same signs โ†’ repulsive (away from each other)
  • Opposite signs โ†’ attractive (toward each other)

Coulomb's Constant

k=14ฯ€ฯต0=8.99ร—109ย N\cdotpm2/C2k = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 \text{ Nยทm}^2\text{/C}^2

where ฯต0=8.85ร—10โˆ’12\epsilon_0 = 8.85 \times 10^{-12} Cยฒ/(Nยทmยฒ) is the permittivity of free space.

Often approximated as: kโ‰ˆ9.0ร—109k \approx 9.0 \times 10^9 Nยทmยฒ/Cยฒ


Comparison with Gravity

Similarities:

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

Differences:

| Property | Gravity | Electricity | |----------|---------|-------------| | Always | Attractive | Attractive OR repulsive | | Constant | G=6.67ร—10โˆ’11G = 6.67 \times 10^{-11} | k=8.99ร—109k = 8.99 \times 10^9 | | Relative strength | Very weak | Much stronger | | Shielding | No | Yes (conductors) |

Electric force is ~10ยณโน times stronger than gravity!


Superposition Principle

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

Fโƒ—total=Fโƒ—1+Fโƒ—2+Fโƒ—3+...\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 โ†’ repel
    • Unlike charges โ†’ attract
  5. Use superposition for multiple charges
  6. Add as vectors (use components if needed)

Common Mistakes

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

๐Ÿ“š Practice Problems

1Problem 1easy

โ“ Question:

Two point charges, qโ‚ = +3.0 ฮผC and qโ‚‚ = -2.0 ฮผC, are separated by 0.50 m. What is the magnitude of the electrostatic force between them? (k = 9.0 ร— 10โน Nยทmยฒ/Cยฒ)

๐Ÿ’ก Show Solution

Given:

  • q1=+3.0q_1 = +3.0 ฮผC =3.0ร—10โˆ’6= 3.0 \times 10^{-6} C
  • q2=โˆ’2.0q_2 = -2.0 ฮผC =โˆ’2.0ร—10โˆ’6= -2.0 \times 10^{-6} C
  • r=0.50r = 0.50 m
  • k=9.0ร—109k = 9.0 \times 10^9 Nยทmยฒ/Cยฒ

Find: Force FF

Solution:

Apply Coulomb's Law: F=kโˆฃq1q2โˆฃr2F = k\frac{|q_1 q_2|}{r^2} F=(9.0ร—109)โˆฃ(3.0ร—10โˆ’6)(โˆ’2.0ร—10โˆ’6)โˆฃ(0.50)2F = (9.0 \times 10^9)\frac{|(3.0 \times 10^{-6})(-2.0 \times 10^{-6})|}{(0.50)^2} F=(9.0ร—109)6.0ร—10โˆ’120.25F = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.25} F=(9.0ร—109)(2.4ร—10โˆ’11)F = (9.0 \times 10^9)(2.4 \times 10^{-11}) F=0.216ย Nโ‰ˆ0.22ย NF = 0.216 \text{ N} \approx 0.22 \text{ N}

Direction: Opposite signs โ†’ attractive force

Answer: 0.22 N, attractive

2Problem 2medium

โ“ Question:

Three charges are arranged in a line: qโ‚ = +2.0 ฮผC at x = 0, qโ‚‚ = -3.0 ฮผC at x = 0.30 m, and qโ‚ƒ = +1.0 ฮผC at x = 0.60 m. What is the net force on qโ‚‚?

๐Ÿ’ก Show Solution

Given:

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

Find: Net force on q2q_2

Solution:

Force from qโ‚ on qโ‚‚:

Distance: r12=0.30r_{12} = 0.30 m F12=kโˆฃq1q2โˆฃr122=(9.0ร—109)(2.0ร—10โˆ’6)(3.0ร—10โˆ’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=(9.0ร—109)6.0ร—10โˆ’120.09=0.60ย NF_{12} = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.09} = 0.60 \text{ N}

Direction: Opposite signs โ†’ attractive โ†’ toward qโ‚ (left, negative) Fโƒ—12=โˆ’0.60ย N\vec{F}_{12} = -0.60 \text{ N}

Force from qโ‚ƒ on qโ‚‚:

Distance: r23=0.60โˆ’0.30=0.30r_{23} = 0.60 - 0.30 = 0.30 m F23=kโˆฃq2q3โˆฃr232=(9.0ร—109)(3.0ร—10โˆ’6)(1.0ร—10โˆ’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=(9.0ร—109)3.0ร—10โˆ’120.09=0.30ย NF_{23} = (9.0 \times 10^9)\frac{3.0 \times 10^{-12}}{0.09} = 0.30 \text{ N}

Direction: Opposite signs โ†’ attractive โ†’ toward qโ‚ƒ (right, positive) Fโƒ—23=+0.30ย N\vec{F}_{23} = +0.30 \text{ N}

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

Answer: 0.30 N to the left (toward qโ‚)

3Problem 3hard

โ“ Question:

Four equal charges Q = +5.0 ฮผC 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?

๐Ÿ’ก Show Solution

Given:

  • Four charges: Q=5.0ร—10โˆ’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=(9.0ร—109)(5.0ร—10โˆ’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=(9.0ร—109)2.5ร—10โˆ’110.04=5.625ย NF_{adj} = (9.0 \times 10^9)\frac{2.5 \times 10^{-11}}{0.04} = 5.625 \text{ N}

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

Force from diagonal charge: Fdiag=kQ2(a2)2=kQ22a2=Fadj2=2.813ย 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: Fโƒ—1=(5.625,0)\vec{F}_1 = (5.625, 0) N From top: Fโƒ—2=(0,5.625)\vec{F}_2 = (0, 5.625) N From diagonal: At 45ยฐ, points toward (along diagonal from origin) $$\vec{F}_3 = (2.813\cos 45ยฐ, 2.813\sin 45ยฐ) = (1.99, 1.99)$ N

Total: Fx=5.625+0+1.99=7.62ย NF_x = 5.625 + 0 + 1.99 = 7.62 \text{ N} Fy=0+5.625+1.99=7.62ย NF_y = 0 + 5.625 + 1.99 = 7.62 \text{ N}

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

ฮธ=tanโกโˆ’1FyFx=45ยฐ\theta = \tan^{-1}\frac{F_y}{F_x} = 45ยฐ

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

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