(a) (i) State Coulomb's law in electrostatics and write it in vector form for two charges. (ii) Gauss's law is

Question

Figure — Figure — CBSE 2023 55/1/1 Q31 — FigureFigure — CBSE 2023 55/1/1 Q31

Q. (a)

(i) State Coulomb's law in electrostatics and write it in vector form for two charges.

(ii) Gauss's law is based on the inverse-square dependence on distance contained in Coulomb's law. Explain.

(iii) Two charges A (charge

q

) and B (charge

2q

) are located at points

(0, 0)

and

(a, a)

respectively. Let

\hat{i}

and

\hat{j}

be the unit vectors along the x-axis and y-axis respectively. Find the force exerted by A on B, in terms of

\hat{i}

and

\hat{j}

(OR)

(b)

(i) Derive an expression for the electric field at a point on the equatorial plane of an electric dipole consisting of charges

q

and

-q

separated by a distance

2a

(ii) The distance of a far-off point on the equatorial plane of an electric dipole is halved. How will the electric field be affected for the dipole?

(iii) Two identical electric dipoles are placed along the diagonals of a square ABCD of side

\sqrt{2}

m as shown in the figure. Obtain the magnitude and direction of the net electric field at the centre (O) of the square.

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Part (a)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.

--- …

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.

--- …

Part (b)Concept understanding — Coulomb Force Superposition

Coulomb Force Superposition – From Intuition to Precision

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)

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

Important

--- …

Why this formula?

Coulomb Force Superposition — Why the Formula Holds

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)

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)

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: $\vec F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}\hat r$ ; Gauss's law follows from the $1/r^2$ law; the force on B is $\vec F_{AB}=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{\sqrt2\,a^2}(\hat i+\hat j)$ . (b) Equatorial field $E=\frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}}\to\frac{p}{4\pi\varepsilon_0 r^3}$ ; halving $r$ multiplies $E$ by 8; the two dipoles give $E_{net}=2\sqrt2\,\frac{q}{4\pi\varepsilon_0}$ at the centre.

Part (a)

(i) Coulomb's law

The force between point charges is proportional to the product of the charges and inversely proportional to the square of their separation, directed along the line joining them:

\vec F_{21}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{|\vec r_2-\vec r_1|^{3}}(\vec r_2-\vec r_1)=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}\hat r_{12}.

(ii) Why Gauss's law needs the inverse-square law

A point charge's field is $E=\dfrac{q}{4\pi\varepsilon_0 r^2}$ . The flux through a concentric sphere of radius $r$ is

\Phi=E\times4\pi r^2=\frac{q}{4\pi\varepsilon_0 r^2}\times4\pi r^2=\frac{q}{\varepsilon_0},

which is independent of $r$ only because the field's $1/r^2$ exactly cancels the surface's $r^2$ . This $r$ -independence (extended to any closed surface via solid angles) is Gauss's law. If the force fell off as any other power, the flux would depend on the surface, and Gauss's law in this simple form would fail.

(iii) Force of A on B

$\vec r_{AB}=a\hat i+a\hat j$ , so $r=a\sqrt2$ and $\hat r_{AB}=\dfrac{1}{\sqrt2}(\hat i+\hat j)$ . With $r^2=2a^2$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 40 on this concept.

CBSE 2026Set A1 markMCQ
$Q. Coulomb's law is valid for (A) Point charges only (B) Dispersed charges only (C) Both point charges and dispersed charges (D) Neutral particles$
›Reveal solutionSolution

Coulomb's law is defined for point charges; extended bodies need integration.

Coulomb's law states $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$ , where $r$ is the distance between the charges.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Two spheres carrying charges +6 μC and +9 μC, separated by a distance d, experience a force of repulsion F. When a charge of -3 μC is added to each sphere and distance d is kept the same, the new force of repulsion will be$
(a) 3F
(b) F/9
(c) F
(d) F/3
›Reveal solutionSolution

New force = F/3, because force is proportional to the product of the charges and only the charges changed, not the distance.

By Coulomb's law, the force between two point charges $q_1$ and $q_2$ separated by a fixed distance $d$ is

$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{d^2}$

Originally $q_1 = +6\ \mu C$ and $q_2 = +9\ \mu C$ , so $F \propto q_1 q_2 = 54$ (in $\mu C^2$ ).

After $-3\ \mu C$ is added to each sphere: …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Two sphere of charge 2μc and 3μc are located at a distance 20 cm apart in air. The ratio of magnitude of electric forces acting between these spheres will be$
(a) 1 : 1
(b) 2 : 3
(c) 3 : 2
(d) 4 : 9
›Reveal solutionSolution

The mutual electric force between two charges is an action-reaction pair, so both spheres feel equal magnitude forces regardless of the charge values.

By Coulomb's law the force sphere 1 exerts on sphere 2 has magnitude F = k q1 q2 / r^2, and the force sphere 2 exerts on sphere 1 has the same magnitude k q1 q2 / r^2, just opposite in direction (Newton's third law applies to electrostatic forces just as it do …

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CBSE 2025Set X11 markMCQ
$q_1$
(a) increases
(b) decreases
(c) may increase or decrease
(d) does not change
›Reveal solutionSolution

(d) does not change. By the principle of superposition, the electrostatic force between $q_1$ and $q_2$ is given by Coulomb's law $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$ and depends …

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CBSE 2025Set D1 markMCQ
$Q. The distance between two charges is made half and one of the charges is also halved. The force acting between the two will become as compared to previous value (A) half (B) double (C) thrice (D) none of these$
›Reveal solutionSolution

Coulomb force F ∝ q₁q₂/r²; halving one charge (×½) and halving the distance (×4) gives a net factor of 2, so the force doubles.

Coulomb's law:

$F = \frac{k q_1 q_2}{r^2}$

Initial force: $F = \dfrac{k q_1 q_2}{r^2}$ .

Now one charge becomes $q_1/2$ and the distance becomes $r/2$ :

…

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CBSE 2025Set D1 markMCQ
$Q. On inserting a dielectric material between two positive charges in air, the value of repulsive force will (A) increase (B) decrease (C) remain same (D) become zero$
›Reveal solutionSolution

A dielectric weakens the field between charges by a factor K, so the repulsive force falls to F₀/K.

The Coulomb force between two charges in air is F₀ = (1/4πε₀)·q₁q₂/r². Filling the space with a dielectric of relative permittivity K replaces ε₀ with Kε₀:

F = (1/4πKε₀)·q₁q₂/r² = F₀/K

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The law governing the force between static electric charges is known as$
(i) Ampere's law
(ii) Ohm's law
(iii) Faraday's law
(iv) Coulomb's law
›Reveal solutionSolution

The force between two static (point) electric charges is governed by Coulomb's law.

Ampere's law relates a magnetic field to the current producing it, Ohm's law relates current and voltage in a conductor, and Faraday's law deals with electromagnetic induction. None of these describes the force between charges at rest.

…

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CBSE 2024Set IMPROVEMENT1 markMCQ
$Q. On placing dielectric material between two point charges in air, repulsive force between them will —$
(a) Increase
(b) Decrease
(c) Remain same
(d) Zero
›Reveal solutionSolution

Placing a dielectric between two charges reduces the force between them.

…

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CBSE 2024Set FS1 markMCQ
$Q. Force of 80 Newton works between two point charges placed at a fixed distance apart in air. When these charges are placed at the same distance apart in a dielectric medium, then force of 8 Newton works on it. The dielectric constant of medium will be:$
(i) $K = -10$
(ii) $K = 10$
(iii) $K = 0.01$
(iv) $K = -0.01$
›Reveal solutionSolution

$K=\dfrac{F_{air}}{F_{medium}}=\dfrac{80}{8}=10$ — option (ii).

Concept. Coulomb's force between two charges at separation $r$ is

$F_{air}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2},\qquad F_{medium}=\frac{1}{4\pi\varepsilon_0 K}\frac{q_1q_2}{r^2}.$

Placing a dielectric of constant $K$ reduces the force by the factor $K$ . …

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CBSE 2024Set A1 mark
$Q. Match Column 'A' with Column 'B' and write the correct pair. Column 'A' item: 'Electrostatic force'. Column 'B' options:$
(i) De-Broglie
(ii) Maxwell
(iii) Ohm
(iv) Einstein
(v) Coulomb
(vi) Lenz
(vii) Young.
›Reveal solutionSolution

Electrostatic force is governed by Coulomb's law.

The force of attraction or repulsion between two stationary point charges is called the electrostatic (or Coulomb) force, and its magnitude is given by Coulomb's law:

$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Two charged spheres are separated by a distance d, exert a force F on each other. If the charges are doubled and the distance between them is doubled then the force is$
(a) F
(b) F/2
(c) F/4
(d) 4F
›Reveal solutionSolution

Coulomb's law force scales as (charge product)/(distance)^2; doubling both charges and the distance leaves the force unchanged.

By Coulomb's law, the force between two point charges q1 and q2 separated by distance d is

$F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{d^2} = k\dfrac{q_1q_2}{d^2}$

…

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CBSE 2024Set ANNUAL1 mark
$Q. What is the name of the electrical force acting between two charges at rest?$
›Reveal solutionSolution

The force between two charges at rest is the electrostatic (Coulomb) force.

The electrical force acting between two charges that are at rest (not moving) is called the electrostatic force, or Coulomb force, since it is governed by Coulomb's law:

$F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r^2}$

…

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Q.(a)

Part (a)

(i) Coulomb's law

(ii) Why Gauss's law needs the inverse-square law

(iii) Force of A on B

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