Coulomb's Law Notes

Coulomb’s Law

• 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: $F = k\frac{|q<em>1||q</em>2|}{r^2}$ 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.

Electrostatics

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

Electric Charges

• 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

Sample Problems

Problem 1

• Three point charges located at the corners of a right triangle, where $q<em>1 = q</em>3 = 5μC$, $q<em>2 = -2μC$, and $a = 0.100 m$. Find the resultant force exerted on $q</em>3$.

• Solution:

  • $F<em>{13}$: Force exerted by $q</em>1$ on $q_3$, repulsive (both charges are positive).

  • $F<em>{23}$: Force exerted by $q</em>2$ on $q_3$, attractive (opposite signs).

• Calculations:

  • $F<em>{13} = k\frac{|q</em>1||q<em>3|}{d</em>{13}^2} = (9\times10^9 \frac{N \cdot m^2}{C^2})\frac{(5\times10^{-6} C)(5\times10^{-6}C)}{(0.100\sqrt{2})^2} = 11.25 N$

  • $F<em>{23} = k\frac{|q</em>2||q<em>3|}{d</em>{13}^2} = (9\times10^9 \frac{N \cdot m^2}{C^2})\frac{(2\times10^{-6} C)(5\times10^{-6}C)}{(0.100)^2} = 9 N$

  • Net force/resultant force at $q_3$:

    • $F = \sqrt{\sum F<em>x^2 + \sum F</em>y^2}$

    • $\sum F<em>x = F</em>{13}\cos 45° - F_{23} = 11.25\cos 45° - 9 = -1.0450 N$

    • $\sum F<em>y = F</em>{13}\sin 45° = 11.25\sin 45° = 7.9550 N$

    • $F_3 = \sqrt{(-1.0450 N)^2 + (7.9550 N)^2} = 8.0233 N$

  • Direction of the resultant force:

    • $\theta = \tan^{-1} \left| \frac{\sum F<em>y}{\sum F</em>x} \right| = \tan^{-1} \left| \frac{7.9550}{1.0450} \right| = 82.5463°$

Problem 2

• Two small beads with positive charges $q<em>1 = 3q$ and $q</em>2 = q$ 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, $F<em>{13} = F</em>{23}$

  • $k\frac{|q<em>1||q</em>3|}{d<em>{13}^2} = k\frac{|q</em>2||q<em>3|}{d</em>{13}^2}$

  • $\frac{3}{x^2} = \frac{1}{(1.50-x)^2}$

  • Solving for $x$ gives $x = 0.9510 m$

Problem 3

• 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, $\sum F<em>x = 0$ and $\sum F</em>y = 0$

  • $\sum F<em>x = T\cos 85° - F</em>e = 0$

  • $k\frac{q^2}{(2L\sin \theta)^2} = T\cos 85°$

  • $L = \sqrt{\frac{kq^2}{(T\cos 85°)(2\sin \theta)^2}}$

  • $L = \sqrt{\frac{(9\times10^9 \frac{N \cdot m^2}{C^2})(7.2\times10^{-9}C)^2}{(1.9695 \times 10^{-3}\cos 85°)(2\sin 5°)^2}} = 0.2991 m$

  • $\sum F_y = T\sin 85° - mg = 0$

  • $T = \frac{mg}{\sin 85°} = \frac{(0.200\times10^{-3}kg)(9.81\frac{m}{s^2})}{\sin 85°} = 1.9695 \times 10^{-3} N$