D2-1. Electric Fields & Coulomb's Law

1. Coulomb's Law & Electric Force

The Law

$$F = k \frac{q_1 q_2}{r^2} \qquad k = \frac{1}{4\pi\varepsilon_0}$$
  • Calculates the electrostatic force between any two point charges.
  • Direction: Attractive for unlike charges; repulsive for like charges.
  • Newton's 3rd Law: The forces on each charge are always equal in magnitude and opposite in direction ($F = F'$).

Notation Key & Units:

  • $F$ = Electrostatic force ($\text{N}$)
  • $k$ = Coulomb constant ($\approx 8.99 \times 10^9\text{ N m}^2\text{ C}^{-2}$)
  • $\varepsilon_0$ = Permittivity of free space
  • $q_1, q_2$ = Interacting point charges ($\text{C}$)
  • $r$ = Separation distance between charges ($\text{m}$)
Repulsion (Like Charges) + q₁ F' + q₂ F r Attraction (Unlike Charges) + q₁ F' - q₂ F r

2. Examples

Example 1

Problem: An electron is placed at a distance of $0.40\text{ m}$ from a fixed point charge of $-6.0\text{ mC}$.

- Fixed Charge -6.0 mC - Electron e⁻ 0.40 m

(a) Show that the electric field strength due to the point charge at the position of the electron is $3.4 \times 10^8\text{ N C}^{-1}$.

(b) (i) Calculate the magnitude of the initial acceleration of the electron.
(ii) Describe the subsequent motion of the electron.

Solution:

(a) The electric field strength is $$ E = k \frac{Q}{r^2} $$ Convert the unit of the charge: $Q = 6.0\text{ mC} = 6.0 \times 10^{-3}\text{ C}$. Then, $$ E = \frac{(8.99 \times 10^9) \times (6.0 \times 10^{-3})}{(0.40)^2} = \frac{5.394 \times 10^7}{0.16} = \mathbf{3.37 \times 10^8\text{ N C}^{-1}} $$ which rounds to $3.4 \times 10^8\text{ N C}^{-1}$.

(b) (i) First, find the force on the electron: $F = qE$. $$ F = (1.60 \times 10^{-19}\text{ C}) \times (3.37 \times 10^8\text{ N C}^{-1}) = 5.39 \times 10^{-11}\text{ N} $$ Now use Newton's Second Law $F = ma$ with the mass of an electron $m_e = 9.11 \times 10^{-31}\text{ kg}$ $$ a = \frac{F}{m} = \frac{5.39 \times 10^{-11}}{9.11 \times 10^{-31}} = \mathbf{5.9 \times 10^{19}\text{ m s}^{-2}} $$

(b) (ii) Both the electron and the point charge are negative, so the electron will be repelled. As the distance $r$ increases, the repulsive force decreases ($F \propto 1/r^2$). Therefore, the electron travels with increasing velocity but decreasing acceleration.

Example 2

Problem: A simple model of the hydrogen atom suggests that the electron orbits the proton. What is the force that keeps the electron in orbit?
A. Electrostatic
B. Gravitational
C. Strong nuclear
D. Centripetal

Solution:

"Centripetal force" is just a mathematical requirement for circular motion, not a fundamental physical force itself. The actual physical force providing this inward pull between the negatively charged electron and positively charged proton is the Electrostatic force. Therefore, the answer is A.

Example 3

Problem: $P$ and $Q$ are two opposite point charges. The force $F$ acting on $P$ due to $Q$ and the electric field strength $E$ at $P$ are shown. Which diagram shows the force on $Q$ due to $P$ and the electric field strength at $Q$?

QUESTION DIAGRAM P Q E F FOUR POSSIBLE FIELD ARROWS ON Q A. Q E F B. Q E F C. Q E F D. Q E F

Solution:

Step 1: The force on $P$ points to the right (towards $Q$). This means they are attracting each other. By Newton's Third Law, the force on $Q$ must point to the left (towards $P$).
Step 2: The electric field at $P$ also points to the right. Since $\vec{F}$ and $\vec{E}$ are in the exact same direction at point $P$, charge $P$ must be positive. Because they are opposite charges, $Q$ must be negative.
Step 3: What is the electric field at $Q$ caused by $P$? Since $P$ is positive, its electric field points away from it. Therefore, at point $Q$, the electric field $\vec{E}$ from $P$ points to the right.
Conclusion: The force on $Q$ is to the Left, and the Field at $Q$ is to the Right.

Example 4

Problem: An electron is held close to the surface of a negatively charged sphere and then released. Which describes the velocity and the acceleration of the electron after it is released?

Velocity Acceleration
A. decreasing constant
B. decreasing decreasing
C. increasing constant
D. increasing decreasing

Solution:

An electron (negative) is repelled by a negative sphere. Because a force is acting on it, it will accelerate, meaning its velocity is increasing. However, as it moves further away, the repulsive electrostatic force gets weaker ($F \propto 1/r^2$). Since force decreases, the acceleration is decreasing. The correct answer is D.

Example 5

Problem: Three fixed charges, $+Q$, $-Q$, and $-2Q$, are at the vertices of an equilateral triangle. What is the resultant force on an electron at the center of the triangle?

+Q +Q -2Q D C B A

An electron has a negative charge. We must evaluate the electrostatic force from each vertex on the central negative electron:

  1. The top-left $+Q$ charge attracts the electron (pulling it up and to the left).
  2. The top-right $+Q$ charge also attracts the electron (pulling it up and to the right).
  3. The bottom $-2Q$ charge strongly repels the electron (pushing it straight up, away from the bottom vertex).

Conclusion: The leftward horizontal pull from the top-left $+Q$ perfectly cancels out the rightward horizontal pull from the top-right $+Q$. However, the vertical components from all three charges point in the exact same direction (upward). When you sum these vector forces, the net resultant force points straight up (North), which corresponds to Arrow D.