(a) (i) State Coulomb's Law in electrostatics. (ii) State the differences between Coulomb force and Gravitatio

Question 107 of 122

Q. (a)

(i) State Coulomb's Law in electrostatics.

(ii) State the differences between Coulomb force and Gravitational force. OR

(b) Describe the Fizeau's method to determine the speed of light.

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Concept understanding — Coulomb Force Superposition

Coulomb Force Superposition – From Intuition to Precision

Imagine you're in a room with three friends. Each friend can push or pull you. If two friends push you from the same side, you feel a stronger push — the combined effect. If one pushes from the left and another from the right, you feel the net effect, which might be smaller or even zero if they push equally hard.

This is exactly how electric forces work. When multiple charged particles are present, each one exerts its own force on a given charge. The total force that charge feels is simply the vector sum of all the individual forces — as if each other charge were acting alone, completely ignoring the presence of the rest.

That's the core idea: forces add like arrows, not like numbers.

The Precise Statement

\vec{F}_{\text{net on } q_0} = \sum_{i=1}^{n} \vec{F}_{i \to 0} = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_0 q_i}{r_{i0}^2} \, \hat{r}_{i0}

Where:

$q_0$ is the charge you're calculating the force on
$q_i$ are all other charges (excluding $q_0$ itself)
$r_{i0}$ is the distance between $q_i$ and $q_0$
$\hat{r}_{i0}$ is a unit vector pointing from $q_i$ to $q_0$ (or away, depending on sign convention — be consistent)

The key point: Each pair of charges interacts independently. The presence of a third charge does not alter the force between the first two. This is what "superposition" means — the forces simply layer on top of each other.

Why This Matters (and a Common Trap)

Watch out

Never add the magnitudes of forces directly unless all forces are along the same line and in the same direction. Force is a vector — direction matters.

If two forces point in opposite directions, they partially cancel. If they're at right angles, the net force is found using the Pythagorean theorem, not simple addition.

Example: Three charges on a line:

$q_1 = +2\,\mu\text{C}$ at $x=0$
$q_2 = -1\,\mu\text{C}$ at $x=3\,\text{cm}$
$q_0 = +1\,\mu\text{C}$ at $x=1\,\text{cm}$

Step 1: Force from $q_1$ on $q_0$ — both positive, so repulsive. $q_0$ is pushed to the right.

Step 2: Force from $q_2$ on $q_0$ — opposite signs, so attractive. $q_0$ is pulled to the right (toward $q_2$ ).

Step 3: Both forces point right. Now you add magnitudes: $F_{\text{net}} = F_{1\to0} + F_{2\to0}$ .

If $q_2$ were also positive, the force from $q_2$ would push $q_0$ left, and you'd subtract.

The Deeper Reason

Coulomb's law is a linear law — the force is proportional to each charge individually. If you double $q_1$ , the force from $q_1$ doubles, but the force from $q_2$ stays the same. This linearity is what makes superposition possible. It's not a coincidence — it's a fundamental property of electromagnetic interactions at the classical level.

Important

Superposition works because electric forces obey a linear inverse-square law. If the force depended on products of three charges (like $q_0 q_1 q_2$ ), superposition would fail. It doesn't — and that's why we can break down any multi-charge problem into a series of two-charge calculations.

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Why this formula?

Coulomb Force Superposition — Why the Formula Holds

The principle of superposition for Coulomb forces states that the net electrostatic force on a given charge due to a collection of other charges is the vector sum of the individual forces from each charge, as if the others were absent.

The Key Formula

If we have a charge $q_0$ at position $\vec{r}_0$ , and $N$ other point charges $q_1, q_2, \dots, q_N$ at positions $\vec{r}_1, \vec{r}_2, \dots, \vec{r}_N$ , the net force on $q_0$ is:

\vec{F}_{\text{net}} = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{N} \frac{q_0 q_i}{|\vec{r}_0 - \vec{r}_i|^2} \, \hat{r}_{0i}

where $\hat{r}_{0i}$ is the unit vector pointing from $q_i$ to $q_0$ .

Why This Works — The Physical Reasoning

1. Coulomb's Law is a Two-Body Interaction

Coulomb's law describes the force between exactly two point charges. It depends only on:

The product of their charges ( $q_0 q_i$ )
The inverse square of the distance between them
The direction along the line joining them

Crucially, the force between $q_0$ and $q_i$ does not depend on the presence of any other charges $q_j$ .

2. Forces Add as Vectors (Newton's Third Law + Linearity)

Electrostatic forces are real physical forces — they obey Newton's laws. If multiple forces act on the same charge, the net effect is the vector sum of each individual force. This is a fundamental property of forces in classical mechanics.

3. The Electric Field is Linear

A deeper reason: the electric field $\vec{E}$ obeys superposition. Since $\vec{F} = q_0 \vec{E}$ , and $\vec{E}$ from multiple sources adds linearly, the force automatically adds linearly.

The electric field at $\vec{r}_0$ due to $q_i$ is:

\vec{E}_i(\vec{r}_0) = \frac{1}{4\pi\varepsilon_0} \frac{q_i}{|\vec{r}_0 - \vec{r}_i|^2} \, \hat{r}_{0i}

Then:

\vec{F}_{\text{net}} = q_0 \sum_i \vec{E}_i = \sum_i \vec{F}_i

The Crucial Assumption (Why It's Not Trivial)

Superposition holds because Maxwell's equations are linear in the electric field. If the equations were nonlinear (e.g., if the field depended on $E^2$ ), then the force from two charges together would not be the sum of the individual forces.

In electrostatics, the electric field satisfies:

\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}, \quad \nabla \times \vec{E} = 0

Both equations are linear — if $\vec{E}_1$ and $\vec{E}_2$ are solutions, then $\vec{E}_1 + \vec{E}_2$ is also a solution. This linearity is the mathematical reason superposition works.

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(a) Coulomb's law gives the electrostatic force between charges and differs from gravity in sign, strength, and medium-dependence; (b) Fizeau's rotating-wheel method measures the speed of light via $c=4DNn$ . Both alternatives answered below.

(a)(i) Coulomb's Law

Statement. The electrostatic (Coulomb) force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges, inversely proportional to the square of the distance between them, and acts along the straight line joining the two charges.

Equation.

$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$

where $q_1,q_2$ are the two point charges, $r$ is the distance between them, and $\dfrac{1}{4\pi\varepsilon_0}=k=9\times10^9\,N\,m^2C^{-2}$ (in vacuum/air) is the Coulomb constant. If $q_1q_2>0$ the force is repulsive; if $q_1q_2<0$ it is attractive.

(a)(ii) Coulomb force vs. Gravitational force

Coulomb (electrostatic) force	Gravitational force
Can be attractive or repulsive, depending on the sign of the charges	Always attractive only
Depends on the medium between the charges (via $\varepsilon_0$ /dielectric constant)	Independent of the intervening medium
Governed by Coulomb's constant $k=9\times10^9\,Nm^2C^{-2}$	Governed by gravitational constant $G=6.67\times10^{-11}\,Nm^2kg^{-2}$
Acts between electric charges	Acts between any two masses
Enormously stronger -- e.g., for two protons the Coulomb force is about $10^{36}$ times the gravitational force between them	Comparatively extremely weak
Can be shielded/reduced by intervening conductors	Cannot be shielded

(b) Fizeau's method to determine the speed of light

Apparatus/setup. Fizeau's method (1849) uses a rotating toothed wheel and a distant fixed mirror. Light from a source is reflected by a half-silvered mirror toward a rotating wheel with $N$ equally spaced teeth and $N$ gaps at its rim. The light passes through a gap in the wheel, travels a long distance $D$ to a fixed plane mirror, is reflected straight back, and returns toward the wheel.

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Q.(a)

(a)(i) Coulomb's Law

(a)(ii) Coulomb force vs. Gravitational force

(b) Fizeau's method to determine the speed of light

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