Coulomb's Law Notes

Electrostatic interaction between charged bodies causes phenomena like a balloon sticking to a wall after being rubbed against hair.
Studies the nature of electric charges and their interactions, described by Coulomb's law.

The magnitude of electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Formula: F = k\frac{|q1||q2|}{r^2} Where:
- k = \frac{1}{4πε_o} \cong 9 \times 10^9 \frac{N \cdot m^2}{C^2}
- F: electric force between the charges in Newtons (N)
- q1, q2: magnitudes of the point charges in Coulombs (C)
- r: distance between the centers of the two charges in meters (m)
- \epsilon_o: permittivity of free space, represents how electric fields interact with a vacuum.
Charges of opposite sign attract, and charges of the same sign repel.
Principle of Superposition:
- When two or more charges exert forces simultaneously on another charge, the total force acting on that charge is the vector sum of the individual forces exerted by each charge.

This principle allows for the calculation of net forces in systems with multiple charges, facilitating accurate predictions of charge interactions.

Deals with phenomena due to attractions or repulsions of electric charges that are not moving.

Electric charge: Basic property of matter carried by some elementary particles.
Governs how particles are affected by electric or magnetic fields.
Two kinds of electric charges: positive and negative.
- Negative charges: possessed by electrons.
  - Charge of electron = -1.6 \times 10^{-19} Coulomb.
- Positive charges: possessed by protons.
  - Charge of proton = 1.6 \times 10^{-19} Coulomb.
Charges of the same sign repel; charges with opposite signs attract.
How do objects get charged?
- POLARIZATION
  - separation of opposite charges within an object
- INSULATORS
  - materials where electrons cannot move freely through the material and are bound to atoms
- CONDUCTORS
  - materials where electrons can move freely through the material and are not bound to atoms.
- SEMICONDUCTORS
  - materials where the electrical properties are somewhere between conductors and insulators

Three point charges located at the corners of a right triangle, where q1 = q3 = 5μC, q2 = -2μC, and a = 0.100 m. Find the resultant force exerted on q3.
Solution:
- F{13}: Force exerted by q1 on q_3, repulsive (both charges are positive).
- F{23}: Force exerted by q2 on q_3, attractive (opposite signs).
Calculations:
- F{13} = k\frac{|q1||q3|}{d{13}^2} = (9\times10^9 \frac{N \cdot m^2}{C^2})\frac{(5\times10^{-6} C)(5\times10^{-6}C)}{(0.100\sqrt{2})^2} = 11.25 N
- F{23} = k\frac{|q2||q3|}{d{13}^2} = (9\times10^9 \frac{N \cdot m^2}{C^2})\frac{(2\times10^{-6} C)(5\times10^{-6}C)}{(0.100)^2} = 9 N
- Net force/resultant force at q_3:
  - F = \sqrt{\sum Fx^2 + \sum Fy^2}
  - \sum Fx = F{13}\cos 45° - F_{23} = 11.25\cos 45° - 9 = -1.0450 N
  - \sum Fy = F{13}\sin 45° = 11.25\sin 45° = 7.9550 N
  - F_3 = \sqrt{(-1.0450 N)^2 + (7.9550 N)^2} = 8.0233 N
- Direction of the resultant force:
  - \theta = \tan^{-1} \left| \frac{\sum Fy}{\sum Fx} \right| = \tan^{-1} \left| \frac{7.9550}{1.0450} \right| = 82.5463°

Two small beads with positive charges q1 = 3q and q2 = q are fixed at opposite ends of a horizontal insulating rod of length d = 1.50 m. The bead with charge q_1 is at the origin. Find the position x where a third small, charged bead is in equilibrium.
Solution:
- Let the third bead have charge q_3 and be located a distance x from the left end of the rod.
- Since the third bead is at equilibrium, F{13} = F{23}
- k\frac{|q1||q3|}{d{13}^2} = k\frac{|q2||q3|}{d{13}^2}
- \frac{3}{x^2} = \frac{1}{(1.50-x)^2}
- Solving for x gives x = 0.9510 m

Two small metallic spheres, each of mass m = 0.200 g, are suspended as pendulums by light strings of length L. The spheres are given the same electric charge of 7.2 nC, and they come to equilibrium when each string is at an angle of \theta = 5.00° with the vertical. How long are the strings? Find the tension in the string.
Solution:
- Consider the free-body diagram of one of the spheres (left sphere). T is the tension in the string, and F_e is the repulsive electrical force exerted by the other sphere.
- The distance between the two spheres is d = 2L\sin \theta
- At equilibrium, \sum Fx = 0 and \sum Fy = 0
- \sum Fx = T\cos 85° - Fe = 0
- k\frac{q^2}{(2L\sin \theta)^2} = T\cos 85°
- L = \sqrt{\frac{kq^2}{(T\cos 85°)(2\sin \theta)^2}}
- L = \sqrt{\frac{(9\times10^9 \frac{N \cdot m^2}{C^2})(7.2\times10^{-9}C)^2}{(1.9695 \times 10^{-3}\cos 85°)(2\sin 5°)^2}} = 0.2991 m
- \sum F_y = T\sin 85° - mg = 0
- T = \frac{mg}{\sin 85°} = \frac{(0.200\times10^{-3}kg)(9.81\frac{m}{s^2})}{\sin 85°} = 1.9695 \times 10^{-3} N