Five identical charges $Q$ are placed equidistant on a semicircle of radius $R$, as shown in the figure. Another point charge $q$ is kept at the centre of the circle. Calculate the electrostatic force experienced by the charge $q$. ![five identical charges Q placed equidistant on a semicircle of radius R with a charge q at the centre — Class 12 Physics electrostatics question](/figures/tn-phy-electrostatics/five-charges-q-equidistant-semicircle-radius-r-q-at-centre-fig-pyq-7bnld3.svg)

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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.

How to Use It in Exams

Draw all charges and label distances.
For each other charge, sketch the direction of the force on your target charge (like charges repel, opposites attract).
Write the magnitude of each force using Coulomb's law.
Resolve into components if forces aren't along the same line.
Add components separately: $F_{\text{net},x} = \sum F_{i,x}$ , same for $y$ , $z$ .
Combine components to get the net force vector.

Tip

In symmetric arrangements (e.g., an equilateral triangle with equal charges), many components cancel. Always check for symmetry before diving into heavy algebra — it can save you minutes.

One Last Check

If you place a test charge $q_0$ at a point and there are 10 other charges around it, you calculate 10 separate Coulomb forces and add them as vectors. That's it. No extra physics, no hidden interactions. The universe, at this level, is beautifully simple: each pair talks only to each other, and you just listen to all the conversations at once.

"Coulomb's law superposition principle examples" and "electrostatics class 12 physics important questions" are frequently searched, both grounded in the Electrostatics chapter of the NCERT/CBSE Class 12 Physics curriculum. Multi-charge force problems using superposition are a near-guaranteed topic in JEE Main and NEET.

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.

Exam-Relevant Takeaway

Concept	Why It Holds
Superposition of forces	Coulomb force is a two-body interaction; forces add as vectors
Superposition of fields	Maxwell's equations are linear in $\vec{E}$
Net force formula	$\vec{F}_{\text{net}} = \sum \vec{F}_i$ — vector sum of individual Coulomb forces

Never forget: The unit vector $\hat{r}_{0i}$ points from the source charge to the test charge — this determines the correct direction of each term.

Quick Example (To Cement the "Why")

Suppose $q_0 = +1\,\mu\text{C}$ at the origin, $q_1 = +2\,\mu\text{C}$ at $(1,0)$ , $q_2 = -2\,\mu\text{C}$ at $(0,1)$ .

Force from $q_1$ : repulsive, along $+x$ direction
Force from $q_2$ : attractive, along $+y$ direction

The net force is not just the sum of magnitudes — it's the vector sum:

\vec{F}_{\text{net}} = F_1 \hat{x} + F_2 \hat{y}

This works because the two forces are independent — $q_1$ doesn't "know" about $q_2$ , and vice versa. The superposition principle is simply the statement that this independence holds.

Step 1. Five identical charges $Q$ sit equally spaced on a semicircle of radius $R$ , with charge $q$ at the centre; each individual force on $q$ has magnitude $F=kQq/R^2$ , directed radially from that charge toward (or away from) the centre.

Step 2. By the symmetry of the semicircular arrangement about its central axis, the components of the five individual forces perpendicular to this axis cancel in pairs (the charge at the middle of the semicircle contributes purely along the axis; the two pairs symmetric about the middle contribute equal and opposite perpendicular components that cancel, while their along-axis components add).

Step 3. Adding the along-axis components -- one contribution of magnitude $F$ from the middle charge, and two pairs each contributing $F\cos45^\circ=F/\sqrt2$ from the charges at $\pm45^\circ$ -- gives a total of $F(1+2\times\tfrac{1}{\sqrt2})=F(1+\sqrt2)$ .

Step 4. Substituting $F=kQq/R^2$ , the net force is $\vec{F}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{qQ}{R^2}(1+\sqrt2)\,\hat{\imath}$ , directed along the semicircle's axis of symmetry.

✓Final answer

\vec{F}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{qQ}{R^2}(1+\sqrt2)\,\hat{\imath}\ \text{N}

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.

This form requires the charges to be point charges so that a single unambiguous separation $r$ exists. For a continuous/dispersed charge distribution the distance from each element differs, so one must integrate Coulomb's law over the distribution rather than apply it directly.

✓Final answer
$(A) Point charges only.$
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:

$q_1' = 6 - 3 = 3\ \mu C$ , $q_2' = 9 - 3 = 6\ \mu C$ , so $F' \propto q_1' q_2' = 18$ (in $\mu C^2$ ).

Since $d$ is unchanged,

$\dfrac{F'}{F} = \dfrac{q_1' q_2'}{q_1 q_2} = \dfrac{18}{54} = \dfrac{1}{3}$

So $F' = F/3$ .

✓Final answer
$(d) F/3.$
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 does to mechanical ones). Since q1 q2 and r are common to both expressions, the two force magnitudes are identical no matter what q1 and q2 individually are.

✓Final answer
$(a) 1 : 1.$
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

✓Final answer
$q_1$
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 $r/2$ :

$F' = \frac{k (q_1/2) q_2}{(r/2)^2} = \frac{k q_1 q_2 /2}{r^2/4} = \frac{4}{2}\cdot\frac{k q_1 q_2}{r^2} = 2F$

The force becomes double the previous value.

✓Final answer
$(B) double.$
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

Since K > 1 for any dielectric, F < F₀. The force stays repulsive (both charges positive) but its magnitude decreases.

✓Final answer
$(B) decrease.$
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.

The electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$ — this is Coulomb's law.

✓Final answer
$(iv) Coulomb's law.$
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.

By Coulomb's law in a medium, $F=\dfrac{1}{4\pi\varepsilon_0 K}\dfrac{q_1q_2}{r^2}$ , where $K$ is the dielectric constant of the medium. Since a dielectric has $K>1$ , placing it between the two point charges (in place of air/vacuum, $K=1$ ) reduces the force by a factor of $K$ . This holds for both attractive and repulsive forces, so the repulsive force between the two positive charges will decrease.

✓Final answer
$(b) Decrease$
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 $K$ .

Solve. With the same charges and separation,

$\frac{F_{air}}{F_{medium}}=K\;\Rightarrow\;K=\frac{80}{8}=10.$

✓Final answer
$K=10$
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:

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

This law was formulated by Charles-Augustin de Coulomb, so "Electrostatic force" pairs with option (v) Coulomb.

✓Final answer
$Electrostatic force \to (v) Coulomb .$
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

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

Now the charges are doubled ( $q_1' = 2q_1$ , $q_2' = 2q_2$ ) and the separation is doubled ( $d' = 2d$ ). The new force is

$F' = k\dfrac{(2q_1)(2q_2)}{(2d)^2} = k\dfrac{4q_1q_2}{4d^2} = k\dfrac{q_1q_2}{d^2} = F$

So the force is exactly unchanged.

✓Final answer
$(a) F.$
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:

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

It acts along the line joining the two charges - repulsive for like charges, attractive for unlike charges. (This is distinct from the additional magnetic force that arises only when charges are in relative motion.)

✓Final answer
$The Coulomb force (electrostatic force) between charges at rest.$

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Q.Five identical charges QQQ are placed equidistant on a semicircle of radius RRR, as shown in the figure. Another point charge qqq is kept at the centre of the circle. Calculate the electrostatic force experienced by the charge qqq.

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Q.Five identical charges $Q$ are placed equidistant on a semicircle of radius $R$ , as shown in the figure. Another point charge $q$ is kept at the centre of the circle. Calculate the electrostatic force experienced by the charge $q$ .

Unlock everything free for 14 days