Electric Charge and Coulomb's Law Notes

Electric Charge and Coulomb's Law

Conductors and Insulators

• Conductors:
  • Can be charged.
  • Charges reside on the surface.
  • Charge distribution described per unit length and per unit area.
• Insulators:
  • Contain very few conduction electrons (approximately $1 electron / cm^3$).
  • Examples: plastic and glass.
  • Can be charged per unit volume, per unit area, and per unit length.

Coulomb's Law

• Statement: The force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
• Formula:$F ∝ \frac{q<em>1 q</em>2}{r^2}$$F = K \frac{q<em>1 q</em>2}{r^2}$
  • Where:
    • $F$ is the force between the charges.
    • $q<em>1$ and $q</em>2$ are the magnitudes of the two charges.
    • $r$ is the distance between the charges.
    • $K$ is the constant of proportionality.
• Nature of Forces:
  • Similar charges repel each other.
  • Unlike charges attract each other.

Units of Charge

• SI System:
  • Unit of current: Ampere (A).
  • Unit of charge: Coulomb (C).
  • Coulomb Definition: The quantity of charge that flows across any cross-section of a wire in one second when there is a steady current of one ampere.
  • If $F$ is in Newtons, $q<em>1 = q</em>2 = 1 C$, and $r = 1 m$, then $F = 8.99 \times 10^9 N$.
  • Therefore, $K = 8.99 \times 10^9 N \cdot m^2/C^2 \approx 9 \times 10^9 N \cdot m^2/C^2$.
  • $\epsilon<em>0 = 8.85 \times 10^{-12} C^2/N \cdot m^2$, where $\epsilon</em>0$ is the permittivity constant of free space.
  • $K = \frac{1}{4 \pi \epsilon_0}$
  • $F = \frac{1}{4 \pi \epsilon<em>0} \frac{q</em>1 q_2}{r^2}$
  • Coulomb Definition (alternative): The amount of charge which, when placed at a distance of 1 m from an equal and similar charge, repels it with a force of approximately $9 \times 10^9 N$.
  • Charge of an electron: $q = -e = -1.6 \times 10^{-19} C$.
  • Charge of a proton: $q = +e = +1.6 \times 10^{-19} C$.
  • Number of electrons/protons needed for 1 C of charge: $\frac{1 C}{1.6 \times 10^{-19} C} = 6.24 \times 10^{18}$.
  • Alternative Statement: A Coulomb is equal in absolute value to the negative or positive charge contained in $6.24 \times 10^{18}$ electrons (protons).
  • Electric force is a vector, and the resultant force is the vector sum of the individual forces acting on the charged particles.
• C.G.S. System:
  • Unit of charge: electrostatic unit (esu).
  • esu Definition: The amount of charge which, when placed at a distance of 1 cm from an equal and similar charge, repels it with a force equal to 1 Dyne.
  • $K = 1 dyne \cdot cm^2/(esu)^2$

Example Problems

Problem 1

• Two identical metal balls, each of 10g mass, carry equal positive charges and are suspended from 1-m long strings.
• The balls repel each other and come to equilibrium, making an angle of $37^o$ with the vertical.
• Goal: Calculate the charge on each ball.
• Solution:
  • $T \sin \theta = mg$
  • $T \cos \theta = F_c$
  • $\tan \theta = \frac{mg}{F_c}$
  • $F_c = k \frac{q^2}{r^2}$
  • $\tan \theta = \frac{mgr^2}{kq^2}$
  • $q^2 = \frac{mgr^2}{k \tan \theta}$
  • $\theta = 90^o - 37^o = 53^o$, so $\tan 53^o = 1.327$
  • $r = AB = 2OB = 2CB \cos \theta = 2(1 m)(0.6) = 1.2 m$
  • $q^2 = \frac{(10 \times 10^{-3} kg)(9.8 m/s^2)(1.2 m)^2}{(9 \times 10^9 N \cdot m^2/C^2)(1.327)}$
  • $q^2 = 1.18 \times 10^{-12} C^2$
  • $q = 3.44 \times 10^{-6} C = 3.44 \mu C$

Problem 2

• Two charges $q<em>1$ and $q</em>2$ are held at a fixed distance $d$ apart.
  • (i) Find the electric force that acts on $q<em>1$ if $q</em>1 = q_2 = 20 \mu C$ and $d = 1.5 m$.
  • (ii) A third charge $q<em>3 = 20 \mu C$ is brought in. Find the new electric force on $q</em>1$.
• Solution:
  • (i) $F = K \frac{q<em>1 q</em>2}{d^2}$
    $F = (9 \times 10^9) \frac{(20 \times 10^{-6})^2}{(1.5)^2} = 1.6 N$
  • (ii) After adding $q<em>3$ the net electric field on $q</em>1$ becomes:
    • $F<em>y = F</em>2 + F_3 \cos 60$
    • $F<em>x = F</em>3 \sin 60$

Problem 3

• Three charged particles (q1, q2, q3) lie on a straight line separated by a distance d.
• q1 and q2 are held fixed, and q3 is free to move and is in equilibrium.
• Goal: Find q1 in terms of q2.
• Solution:
  • For q3 to be in equilibrium, $F<em>1 = -F</em>2$.
  • This means if q1 is positive, q2 is negative, and vice versa (q1 and q2 are different charges).
  • $F<em>1 = K \frac{q</em>1 q<em>3}{(2d)^2} = K \frac{q</em>1 q_3}{4d^2}$
  • $F<em>2 = K \frac{q</em>2 q_3}{d^2}$
  • Equating the forces: $K \frac{q<em>1 q</em>3}{4d^2} = K \frac{q<em>2 q</em>3}{d^2}$
  • $q<em>1 = -4q</em>2$ or $q<em>2 = -\frac{q</em>1}{4}$

Problem 4

• Charges q1 and q2 lie on the x-axis at points x = -a and x = +a, respectively.
  • (a) How must q1 and q2 be related for the net force on charge +Q, placed at x = +a/2, to be zero?
  • (b) Answer the same question if the +Q charge is placed at x = +3a/2?

Problem 5

• Find the horizontal and vertical components of the resultant electric force on the charge in the lower-left corner of a square.
• Assume $q = 1.0 \times 10^{-7} C$ and $a = 5.0 cm$.
• The charges are at rest.