Electric force between point charges

1. Overview

The electrostatic force is a fundamental interaction that governs the behavior of all charged matter, from the binding of electrons in atoms to the macroscopic behavior of static electricity. At the A-Level, this interaction is quantified by Coulomb’s Law, which defines the force between two point charges. A critical conceptual bridge in this topic is the point charge approximation, which allows us to apply Coulomb's Law to spherical objects, such as metal spheres or atomic nuclei, by treating their entire charge as if it were concentrated at a single central point. This principle is essential for solving problems involving planetary-scale electrostatic models or subatomic particle interactions.

Key Definitions

• Coulomb’s Law: The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their separation.
• Point Charge: A theoretical model of a charged object where the physical dimensions (size/radius) are negligible compared to the distances between the objects being considered.
• Permittivity of Free Space ($\varepsilon_0$): A physical constant representing the resistance encountered when forming an electric field in a vacuum. Its value is approximately $8.85 \times 10^{-12} \text{ F m}^{-1}$.
• Elementary Charge ($e$): The smallest unit of electric charge found in nature (the charge of a proton or electron), equal to $1.60 \times 10^{-19} \text{ C}$.
• Inverse Square Law: A physical law stating that a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity.

Content

3.1 Spherical Conductors as Point Charges

In many practical physics scenarios, we do not deal with infinitesimal points but with three-dimensional objects like metal spheres. To use Coulomb's Law, we must understand when these objects can be simplified.

• The Principle: For any point outside a uniformly charged spherical conductor, the charge on the sphere may be considered to act as a point charge situated at its geometric centre.
• The Mechanism of Uniformity: On a conducting sphere, like-charges repel each other and move as far apart as possible. This results in the charge distributing itself uniformly over the outer surface.
• Field Symmetry: Because the charge is uniform, the electric field lines outside the sphere are perfectly radial (they point directly away from or toward the centre). If an observer outside the sphere traces these lines backward, they appear to originate from a single point at the centre.
• The Boundary Condition: This approximation is only valid for distances $r \geq R$, where $R$ is the radius of the sphere.
  • If $r > R$, the sphere behaves exactly like a point charge of magnitude $Q$ at the centre.
  • If $r < R$ (inside the conductor), the electric field is zero, and the point charge model is no longer valid.
• Application: When calculating the force between two charged spheres, the distance $r$ in Coulomb's Law must be measured from the centre of one sphere to the centre of the other, not from their surfaces.

3.2 Coulomb’s Law

Coulomb’s Law provides the mathematical description of the force $F$ between two stationary point charges $Q_1$ and $Q_2$ separated by a distance $r$ in a vacuum (free space).

The formula is expressed as: $$\mathbf{F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}}$$

Breakdown of the Components:

1. The Product $Q_1 Q_2$: The force is proportional to the magnitude of both charges. If you double one charge, the force doubles. If you double both, the force quadruples.
2. The Inverse Square $r^2$: The force is highly sensitive to distance. If the separation $r$ is tripled, the force becomes $\frac{1}{3^2} = \frac{1}{9}$ of its original value.
3. The Constant $4\pi\varepsilon_0$:
   • $\varepsilon_0$ is the permittivity of free space.
   • The $4\pi$ factor arises from the three-dimensional geometry of the sphere surrounding the point charge (the surface area of a sphere is $4\pi r^2$).
   • In air, the permittivity is very close to $\varepsilon_0$, so we use the same constant for both vacuum and air calculations.

Force as a Vector:

• Direction: The force acts along the straight line joining the two charges.
• Attraction vs. Repulsion:
  • Like charges (e.g., $+Q$ and $+Q$) result in a positive value for $F$, indicating a repulsive force.
  • Opposite charges (e.g., $+Q$ and $-Q$) result in a negative value for $F$, indicating an attractive force.
• Newton's Third Law: The force exerted by $Q_1$ on $Q_2$ is equal in magnitude and opposite in direction to the force exerted by $Q_2$ on $Q_1$.

3.3 The Proportionality Constant $k$

In some contexts, Coulomb's Law is simplified using the Coulomb constant $k$: $$F = \frac{k Q_1 Q_2}{r^2}$$ Where: $$k = \frac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$$ Note for Exams: While $k$ is convenient, the Cambridge 9702 Data Sheet provides $\varepsilon_0$. It is highly recommended to use the full $\frac{1}{4\pi\varepsilon_0}$ expression in your working to show a clear understanding of the fundamental constants.

3.4 Comparison with Newton’s Law of Gravitation

There is a striking mathematical symmetry between Coulomb’s Law and Newton’s Law of Gravitation ($F = \frac{Gm_1 m_2}{r^2}$).

Feature	Electrostatic Force (Coulomb)	Gravitational Force (Newton)
Equation	$F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}$	$F = \frac{G m_1 m_2}{r^2}$
Dependency	Inverse square of distance ($1/r^2$)	Inverse square of distance ($1/r^2$)
Property	Dependent on Charge	Dependent on Mass
Nature	Can be Attractive or Repulsive	Always Attractive
Medium	Affected by the medium (permittivity)	Independent of the medium
Strength	Very strong (e.g., $k \approx 10^{10}$)	Very weak (e.g., $G \approx 10^{-11}$)

Conceptual Insight: At the atomic level, the gravitational force between a proton and an electron is roughly $10^{39}$ times weaker than the electrostatic force. This is why gravity is ignored in subatomic physics but dominates on a planetary scale where large masses are usually electrically neutral.

---

3.5 Worked Examples

Worked example 1 — Basic Force Calculation

Two small conducting spheres are charged to $+5.0 \text{ \mu C}$ and $-3.0 \text{ \mu C}$ respectively. They are placed so that their centres are $12 \text{ cm}$ apart in a vacuum. Calculate the magnitude of the electrostatic force between them and state its nature.

Step 1: Identify and convert units

• $Q_1 = +5.0 \times 10^{-6} \text{ C}$
• $Q_2 = -3.0 \times 10^{-6} \text{ C}$
• $r = 0.12 \text{ m}$ (Distance must be in metres)
• $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$

Step 2: State the equation $$F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}$$

Step 3: Substitute values (using magnitudes for $F$) $$F = \frac{(5.0 \times 10^{-6}) \times (3.0 \times 10^{-6})}{4 \times \pi \times (8.85 \times 10^{-12}) \times (0.12)^2}$$

Step 4: Intermediate calculation

• Numerator: $1.5 \times 10^{-11}$
• Denominator: $1.601... \times 10^{-12}$

Step 5: Final Answer $$F = 9.366... \text{ N}$$ Answer: $F = 9.4 \text{ N}$ (to 2 s.f.). Since the charges have opposite signs, the force is attractive.

Worked example 2 — Atomic Scale Interactions

In a helium atom, the two electrons can be modeled as being at a distance of $2.6 \times 10^{-11} \text{ m}$ from each other at a specific instant. Calculate the repulsive force between the two electrons.

Step 1: Identify constants

• Charge of an electron $Q_1 = Q_2 = -1.60 \times 10^{-19} \text{ C}$
• $r = 2.6 \times 10^{-11} \text{ m}$

Step 2: Substitute into Coulomb's Law $$F = \frac{(-1.60 \times 10^{-19})^2}{4\pi \times (8.85 \times 10^{-12}) \times (2.6 \times 10^{-11})^2}$$

Step 3: Calculation $$F = \frac{2.56 \times 10^{-38}}{7.51 \times 10^{-32}}$$ $$F = 3.408... \times 10^{-7} \text{ N}$$

Answer: $F = 3.4 \times 10^{-7} \text{ N}$. (Note: While this force seems small, the mass of an electron is so tiny that this force produces an incredible acceleration of $\approx 3.7 \times 10^{23} \text{ m s}^{-2}$).

Worked example 3 — Finding the Point of Zero Force

Two point charges, $A = +4.0 \text{ \mu C}$ and $B = +9.0 \text{ \mu C}$, are fixed $1.0 \text{ m}$ apart. A third charge $q$ is placed on the line joining $A$ and $B$ such that the net electrostatic force on $q$ is zero. Calculate the distance of $q$ from charge $A$.

Step 1: Analysis For the net force to be zero, the force from $A$ ($F_A$) must be equal in magnitude and opposite in direction to the force from $B$ ($F_B$). Let the distance from $A$ be $x$. Therefore, the distance from $B$ is $(1.0 - x)$.

Step 2: Set up the equilibrium equation $$\frac{Q_A q}{4\pi\varepsilon_0 x^2} = \frac{Q_B q}{4\pi\varepsilon_0 (1.0 - x)^2}$$

Step 3: Simplify Notice that $q$ and $4\pi\varepsilon_0$ cancel out on both sides: $$\frac{Q_A}{x^2} = \frac{Q_B}{(1.0 - x)^2}$$

Step 4: Substitute values and solve $$\frac{4.0 \times 10^{-6}}{x^2} = \frac{9.0 \times 10^{-6}}{(1.0 - x)^2}$$ $$\frac{4}{x^2} = \frac{9}{(1.0 - x)^2}$$

Take the square root of both sides: $$\frac{2}{x} = \frac{3}{1.0 - x}$$ $$2(1.0 - x) = 3x$$ $$2.0 - 2x = 3x \implies 5x = 2.0 \implies x = 0.40 \text{ m}$$

Answer: The charge $q$ must be placed $0.40 \text{ m}$ from charge $A$.

---

Key Equations

Equation	Symbols	SI Units	Status
$F = \frac{Q_1 Q_2}{4\pi\varepsilon_0 r^2}$	$F$: Electric force $Q$: Charge $\varepsilon_0$: Permittivity of free space $r$: Separation of centres	$\text{N}$ $\text{C}$ $\text{F m}^{-1}$ $\text{m}$	On Data Sheet
$e = 1.60 \times 10^{-19} \text{ C}$	Elementary charge	$\text{C}$	On Data Sheet
$\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$	Permittivity of free space	$\text{F m}^{-1}$	On Data Sheet
$k = \frac{1}{4\pi\varepsilon_0}$	Coulomb constant ($\approx 8.99 \times 10^9$)	$\text{N m}^2 \text{ C}^{-2}$	Calculate from $\varepsilon_0$

---

Common Mistakes to Avoid

• ❌ Wrong: Using the diameter of a sphere as $r$ or using the distance between the surfaces of two spheres.
• ✅ Right: Always use the distance between the centres of the spheres. If the question says "two spheres of radius $2 \text{ cm}$ are separated by $10 \text{ cm}$ between their surfaces," then $r = 2 + 10 + 2 = 14 \text{ cm}$.
• ❌ Wrong: Forgetting to square the distance $r$ in the denominator.
• ✅ Right: This is the most common calculation error. Always double-check the $r^2$ term in your calculator.
• ❌ Wrong: Using charge values in $\mu\text{C}$, $\text{nC}$, or $\text{pC}$ directly in the formula.
• ✅ Right: Convert all prefixes to base units: $1 \mu\text{C} = 10^{-6} \text{ C}$, $1 \text{ nC} = 10^{-9} \text{ C}$, $1 \text{ pC} = 10^{-12} \text{ C}$.
• ❌ Wrong: Confusing the permittivity of free space ($\varepsilon_0$) with the elementary charge ($e$).
• ✅ Right: Check the Data Sheet carefully. $\varepsilon_0$ is $8.85 \times 10^{-12}$, while $e$ is $1.60 \times 10^{-19}$.
• ❌ Wrong: Assuming the force inside a charged hollow sphere is calculated using the point charge at the centre.
• ✅ Right: The electric field (and thus the force on any charge) inside a uniform spherical conductor is zero. The point charge model only works for points outside.

---

Exam Tips

1. The $1/r^2$ Relationship: If an exam question asks how the force changes when the distance is halved, you don't need to re-calculate everything. Since $F \propto 1/r^2$, if $r \to \frac{1}{2}r$, then $F \to \frac{1}{(1/2)^2}F = 4F$. The force quadruples.
2. Graphing Skills:
   • A graph of $F$ against $r$ is a curve (hyperbola-like) that never touches the axes.
   • A graph of $F$ against $1/r^2$ is a straight line through the origin. The gradient of this line is $\frac{Q_1 Q_2}{4\pi\varepsilon_0}$.
3. Vector Addition: If you are asked to find the resultant force on a charge $C$ due to charges $A$ and $B$, calculate $F_{AC}$ and $F_{BC}$ separately. If they are in a line, add or subtract them based on direction. If they are at an angle, use a vector triangle or resolve into components.
4. Significant Figures: Cambridge 9702 is strict on significant figures. If the data in the question is given to 2 s.f. (e.g., $5.0 \text{ cm}$), provide your final answer to 2 s.f.
5. Ratio Problems: Many questions involve comparing the force between two protons to the force between two alpha particles at the same distance.
   • Proton charge = $+e$
   • Alpha particle charge = $+2e$
   • Force between protons: $F \propto (e)(e) = e^2$
   • Force between alpha particles: $F \propto (2e)(2e) = 4e^2$
   • The force between alpha particles is 4 times greater.