Coulomb’s Law is deceptively simple. The scalar formula $F = k \frac{|q_1 q_2|}{r^2}$ takes five seconds to memorize. But when university physics and electrical engineering exams test this concept, they rarely ask for a simple 1D calculation. They embed the charges in 2D or 3D space, forcing you to manage vector decomposition, unit conversions, and attractive versus repulsive directions.
If you are looking for rigorous Coulomb's law examples that bridge the gap between textbook formulas and actual exam problems, this walkthrough is your blueprint. We will solve a classic 2D right-triangle configuration, show every algebraic step without skipping the powers of ten, and build a sanity check to ensure your final answer is physically sound.
The Decision Tree: Which Method Applies to Your Electrostatics Problem?
Before writing a single formula, you must classify the geometry of the charge distribution. Choosing the wrong mathematical approach is the primary reason students run out of time on exams. Use this decision path to select your method.
| Charge Geometry | Symmetry Present? | Required Method | When to Use |
|---|---|---|---|
| 1D Collinear (on a single axis) | N/A | Scalar Addition | Charges are placed along a single straight line. Assign right/up as positive, left/down as negative. |
| 2D Symmetric (e.g., center of a square, equilateral triangle) | Yes | Symmetry Cancellation | Opposing charges cancel out specific vector components (e.g., $F_x$ cancels, leaving only $F_y$). Do not calculate canceled vectors. |
| 2D Asymmetric (e.g., right triangle, arbitrary coordinates) | No | Vector Decomposition | Calculate magnitude first, resolve into $x$ and $y$ components using $\sin/\cos$, sum the components, then recombine via Pythagorean theorem. |
Practice Problem: 2D Vector Coulomb's Law in a Right Triangle
Three point charges are arranged in a right triangle on an x-y plane:
• Charge $q_1 = +2 \mu C$ is located at the origin $(0, 0)$.
• Charge $q_2 = -3 \mu C$ is located on the x-axis at $(0.03 \text{ m}, 0)$.
• Charge $q_3 = +4 \mu C$ is located on the y-axis at $(0, 0.04 \text{ m})$.
Calculate the magnitude and direction (angle relative to the positive x-axis) of the net electrostatic force acting on $q_1$. Use $k = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$ for exam simplicity.
Step-by-Step Algebraic Solution
We will use the Vector Decomposition method identified in our decision tree. According to Georgia State University's HyperPhysics, the superposition principle dictates that we calculate the force from each charge independently and sum them as vectors.
Step 1: Unit Conversion and Setup
The most common fatal error in Coulomb's law examples is failing to convert microcoulombs ($\mu C$) and centimeters to base SI units.
- $q_1 = +2 \times 10^{-6} \text{ C}$
- $q_2 = -3 \times 10^{-6} \text{ C}$
- $q_3 = +4 \times 10^{-6} \text{ C}$
- $r_{12} = 0.03 \text{ m}$ (Distance from $q_1$ to $q_2$)
- $r_{13} = 0.04 \text{ m}$ (Distance from $q_1$ to $q_3$)
Step 2: Calculate Force Magnitudes
Calculate the scalar magnitude of the force exerted on $q_1$ by $q_2$ ($F_{12}$):
$$F_{12} = k \frac{|q_1 q_2|}{r_{12}^2}$$
$$F_{12} = (9 \times 10^9) \frac{|(2 \times 10^{-6})(-3 \times 10^{-6})|}{(0.03)^2}$$
$$F_{12} = (9 \times 10^9) \frac{6 \times 10^{-12}}{9 \times 10^{-4}}$$
$$F_{12} = \frac{54 \times 10^{-3}}{9 \times 10^{-4}} = 6 \times 10^1 = 60 \text{ N}$$
Calculate the scalar magnitude of the force exerted on $q_1$ by $q_3$ ($F_{13}$):
$$F_{13} = k \frac{|q_1 q_3|}{r_{13}^2}$$
$$F_{13} = (9 \times 10^9) \frac{|(2 \times 10^{-6})(4 \times 10^{-6})|}{(0.04)^2}$$
$$F_{13} = (9 \times 10^9) \frac{8 \times 10^{-12}}{16 \times 10^{-4}}$$
$$F_{13} = \frac{72 \times 10^{-3}}{16 \times 10^{-4}} = 4.5 \times 10^1 = 45 \text{ N}$$
Step 3: Determine Vector Directions
- $F_{12}$ Direction: $q_1$ is positive and $q_2$ is negative. Opposite charges attract. Therefore, $q_2$ pulls $q_1$ in the positive x-direction. Vector: $\vec{F}_{12} = +60\hat{i} \text{ N}$.
- $F_{13}$ Direction: $q_1$ and $q_3$ are both positive. Like charges repel. Therefore, $q_3$ pushes $q_1$ in the negative y-direction. Vector: $\vec{F}_{13} = -45\hat{j} \text{ N}$.
Step 4: Vector Addition and Final Angle
Because the charges form a right triangle aligned with the axes, the vectors are already orthogonal. We skip the $\sin/\cos$ resolution step and apply the Pythagorean theorem directly.
$$|\vec{F}_{net}| = \sqrt{(F_x)^2 + (F_y)^2}$$
$$|\vec{F}_{net}| = \sqrt{(60)^2 + (-45)^2} = \sqrt{3600 + 2025} = \sqrt{5625} = 75 \text{ N}$$
To find the angle $\theta$ relative to the positive x-axis:
$$\theta = \arctan\left(\frac{F_y}{F_x}\right) = \arctan\left(\frac{-45}{60}\right) = \arctan(-0.75) \approx -36.87^\circ$$
Final Answer: The net force is 75 N at an angle of -36.87° (or 323.13° standard position).
The Trap & Sanity Check: Verifying Your Answer
How do you verify this answer independently before moving to the next exam question? You run a two-part sanity check focusing on dimensional analysis and order of magnitude.
The Trap: Forgetting to Square the Distance
The most frequent algebraic error in MIT OpenCourseWare's 8.02 Electricity and Magnetism problem sets is using $r$ instead of $r^2$ in the denominator. If you had forgotten to square $0.03$, your $F_{12}$ would have been $1800 \text{ N}$—a massive red flag. Always write the squared term explicitly in your setup.
Sanity Check 1: Dimensional Analysis (Units)
Verify that your units collapse into Newtons (N). The NIST CODATA value for Coulomb's constant $k$ carries units of $\text{N}\cdot\text{m}^2/\text{C}^2$.
$$\left( \frac{\text{N} \cdot \text{m}^2}{\text{C}^2} \right) \times \left( \frac{\text{C} \cdot \text{C}}{\text{m}^2} \right) = \text{N}$$
The units cancel perfectly. If your final unit isn't Newtons, you missed a unit conversion (usually leaving distance in cm).
Sanity Check 2: Order of Magnitude
Do a quick mental math check of the powers of ten.
- $k$ is on the order of $10^9$.
- Two $\mu C$ charges multiplied together yield $10^{-12}$.
- The numerator is roughly $10^{-3}$.
- Distances in cm squared ($0.03^2$) yield roughly $10^{-3}$ to $10^{-4}$ in the denominator.
- $10^{-3} / 10^{-4} = 10^1$.
Your final answer should be in the tens or hundreds of Newtons. Our answer of 75 N fits this order of magnitude perfectly. If you calculated $0.075 \text{ N}$ or $75,000 \text{ N}$, you dropped a decimal place.
FAQ: Coulomb's Law Exam Gotchas
Q: Why did we use $k = 9 \times 10^9$ instead of the exact $8.987 \times 10^9$?
A: In timed university exams, professors design the numbers (like our 3-4-5 right triangle) to cancel out cleanly with $9 \times 10^9$. Using $8.987$ will yield $74.89 \text{ N}$, which is technically more accurate but will cost you precious minutes and break the integer geometry intended by the test maker. Always check your exam instructions for the preferred constant.
Q: What if the charges are embedded in a material like water or glass?
A: Coulomb's constant $k$ assumes a vacuum (or air, which is close enough for exams). If the problem specifies a dielectric medium, you must divide $k$ by the relative permittivity (dielectric constant, $\kappa$) of that material. The formula becomes $F = \frac{k |q_1 q_2|}{\kappa r^2}$. Water, for example, has a $\kappa$ of roughly 80, which would reduce our 75 N force to less than 1 N.
Q: Should I calculate the force $q_1$ exerts on $q_2$, or $q_2$ exerts on $q_1$?
A: By Newton's Third Law, the magnitudes are exactly identical. However, the question asked for the net force acting on $q_1$. Therefore, all vector arrows in your free-body diagram must originate from $q_1$. Drawing the vectors originating from the source charges is a common graphical trap that leads to flipped signs in your final coordinate summation.
