Two charges q and -3q are placed fixed on x-axis separated by distance d. Where should a third charge 2q be pl

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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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The key idea is the Principle of Superposition of Coulomb Forces.

For the third charge $2q$ to experience no net force, the electrostatic forces exerted on it by $q$ and $-3q$ must be equal in magnitude and opposite in direction.

Identify possible regions: Since $q$ and $-3q$ have opposite signs, the forces on $2q$ will be in opposite directions only if $2q$ is placed outside the region between $q$ and $-3q$ .
Determine specific region: For the forces to cancel, $2q$ must be closer to the charge with the smaller magnitude. Here, $|q| < |-3q|$ , so $2q$ must be placed to the left of $q$ .
Set up coordinates and forces: Let $q$ be at $x=0$ and $-3q$ be at $x=d$ . Let the third charge $2q$ be placed at position $x$ (where $x<0$ ). The force due to $q$ on $2q$ is $F_1 = k \frac{q(2q)}{(-x)^2}$ (repulsive, to the left). The force due to $-3q$ on $2q$ is $F_2 = k \frac{|-3q|(2q)}{(d-x)^2}$ (attractive, to the right).
Equate magnitudes: For zero net force, $|F_1| = |F_2|$ :

k \frac{2q^2}{x^2} = k \frac{6q^2}{(d-x)^2}

\frac{1}{x^2} = \frac{3}{(d-x)^2}

$$(d-x)^2 = 3x^2$$ …

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For the net force on a third charge to be zero, the forces from the other two charges must be equal in magnitude and opposite in direction. This occurs at a point outside the segment connecting the two charges, closer to the charge with the smaller magnitude. The third charge $2q$ should be placed at a distance of $\frac{d(1+\sqrt{3})}{2}$ to the left of the charge $q$ .

When multiple charges exert forces on a single charge, the net force on that charge is the vector sum of all individual forces. This is known as the Principle of Superposition. For a third charge to experience no net force, the individual forces acting on it must perfectly cancel each other out. This means their magnitudes must be equal, and their directions must be opposite.

Let's set up the problem:

We have two fixed charges:

$q_1 = q$
$q_2 = -3q$ They are separated by a distance $d$ . Let's place $q_1$ at the origin ( $x=0$ ) and $q_2$ at $x=d$ on the x-axis. We want to find the position $x$ where a third charge $q_3 = 2q$ experiences zero net force.

Q_A

We need to consider three possible regions along the x-axis where the third charge $q_3$ could be placed:

Region I: To the left of $q_1$ ( $x < 0$ )
Region II: Between $q_1$ and $q_2$ ( $0 < x < d$ )
Region III: To the right of $q_2$ ( $x > d$ )

Let's analyze each region:

1. Analyze the possible regions for zero net force

We need the forces from $q_1$ and $q_2$ on $q_3$ to be equal in magnitude and opposite in direction.

Region II ( $0 < x < d$ ):
- $q_1$ (positive) will repel $q_3$ (positive) to the right.
- $q_2$ (negative) will attract $q_3$ (positive) to the right.
- In this region, both forces act in the same direction (towards the positive x-axis). Therefore, they cannot cancel each other out, and the net force will never be zero. No solution exists in this region.
Region I ( $x < 0$ ) and Region III ( $x > d$ ):
- In these regions, the forces from $q_1$ and $q_2$ on $q_3$ will be in opposite directions. For example, if $q_3$ is to the left of $q_1$ , $q_1$ repels $q_3$ to the left, while $q_2$ attracts $q_3$ to the right. If $q_3$ is to the right of $q_2$ , $q_1$ repels $q_3$ to the right, while $q_2$ attracts $q_3$ to the left. Thus, cancellation is possible in these regions.
Tip
For the forces to cancel, the third charge must be closer to the charge with the smaller magnitude. Here, $|q_1| = q$ and $|q_2| = |-3q| = 3q$ . Since $|q_1| < |q_2|$ , the null point must be closer to $q_1$ .
- In Region I ( $x < 0$ ), $q_3$ is closer to $q_1$ than to $q_2$ . This is consistent with the tip.
- In Region III ( $x > d$ ), $q_3$ is closer to $q_2$ than to $q_1$ . This contradicts the tip, meaning no solution exists in Region III.
Therefore, the only possible region for the null point is Region I ( $x < 0$ ).

2. Set up the force equations for Region I

Let the position of the third charge $q_3 = 2q$ be $x$ , where $x < 0$ .

Force from $q_1$ on $q_3$ ( $F_1$ ):
- Charges $q_1=q$ and $q_3=2q$ are both positive, so the force is repulsive.
- The distance between $q_1$ (at $x=0$ ) and $q_3$ (at $x$ ) is $r_1 = |x| = -x$ (since $x$ is negative).
- The force $F_1$ acts in the negative x-direction (to the left).
- Magnitude: $F_1 = k \frac{|q_1 q_3|}{r_1^2} = k \frac{q(2q)}{(-x)^2} = k \frac{2q^2}{x^2}$
Force from $q_2$ on $q_3$ ( $F_2$ ):
- Charges $q_2=-3q$ and $q_3=2q$ have opposite signs, so the force is attractive.
- The distance between $q_2$ (at $x=d$ ) and $q_3$ (at $x$ ) is $r_2 = |d-x| = d-x$ (since $x < 0$ , $d-x$ is positive).
- The force $F_2$ acts in the positive x-direction (to the right).
- Magnitude: $F_2 = k \frac{|q_2 q_3|}{r_2^2} = k \frac{|-3q|(2q)}{(d-x)^2} = k \frac{6q^2}{(d-x)^2}$

3. Equate the magnitudes of the forces …

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Concept: Superposition of Electrostatic Forces

The net force on a charge due to multiple other charges is the vector sum of the individual Coulomb forces. For equilibrium (zero net force), the forces from $q$ and $-3q$ on $2q$ must cancel each other.

Method: Force Balance with Sign Analysis

Step 1: Understand the force directions

$q$ is positive, $-3q$ is negative.
$2q$ is positive.
Like charges repel, unlike charges attract.

So:

Force from $q$ on $2q$ → repulsive (pushes $2q$ away from $q$ )
Force from $-3q$ on $2q$ → attractive (pulls $2q$ toward $-3q$ )

Step 2: Identify possible regions

Place $q$ at $x=0$ and $-3q$ at $x=d$ .

Region	Force from $q$	Force from $-3q$	Can they cancel?
Left of $q$ ( $x<0$ )	Left (repel)	Left (attract)	No (both same direction)
Between $q$ and $-3q$ ( $0<x<d$ )	Right (repel)	Right (attract)	No (both same direction)
Right of $-3q$ ( $x>d$ )	Right (repel)	Left (attract)	Yes (opposite directions)

Only the region $x > d$ works.

Step 3: Apply Coulomb’s law for equilibrium

Let the third charge be at $x = r$ (with $r > d$ ).

Distance from $q$ (at $x=0$ ) to $2q$ : $r$

Distance from $-3q$ (at $x=d$ ) to $2q$ : $r-d$

Force magnitudes must be equal:

\frac{k \cdot q \cdot 2q}{r^2} = \frac{k \cdot 3q \cdot 2q}{(r-d)^2}

Cancel $k$ , $2q^2$ (both sides):

\frac{1}{r^2} = \frac{3}{(r-d)^2}

Step 4: Solve for $r$

Take square roots (positive, since distances are positive):

\frac{1}{r} = \frac{\sqrt{3}}{r-d}

Cross-multiply:

r-d = \sqrt{3} \, r

-d = (\sqrt{3} - 1) r

r = \frac{-d}{\sqrt{3} - 1}

Since $r$ must be positive, and $\sqrt{3} - 1 > 0$ , the negative sign tells us we took the wrong sign in the square root. Let’s correct:

From $\frac{1}{r^2} = \frac{3}{(r-d)^2}$ , taking square root gives: …

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Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the Sign of the Force

The error: Students treat all forces as repulsive. They forget that $-3q$ is negative, so the force between $2q$ and $-3q$ is attractive, not repulsive.

How to avoid:

Always write the sign of each charge explicitly.
Use the rule:
- Like charges → repulsion (force away from each other)
- Unlike charges → attraction (force toward each other)
Draw arrows on a diagram showing direction of each force on $2q$ .

Mistake 2: Placing the Third Charge Only Between the Two Fixed Charges

The error: Students assume the third charge must lie between $q$ and $-3q$ . They don't check the left side or right side regions.

How to avoid:

The net force is zero when the magnitudes of the two forces are equal and their directions are opposite.
On the $x$ $x$ -axis, there are three possible regions:
1. Left of $q$ ( $x < 0$ )
2. Between $q$ and $-3q$ ( $0 < x < d$ )
3. Right of $-3q$ ( $x > d$ )
Test each region systematically. For this problem, the correct location is left of $q$ .

Mistake 3: Using Wrong Distance in Coulomb’s Law

The error: Students take $d$ as the distance from $2q$ to both charges, or use $d$ incorrectly when the third charge is not between them.

How to avoid:

Let the position of $2q$ be at coordinate $x$ (with $q$ at $x=0$ , $-3q$ at $x=d$ ).
Distance to $q$ : $|x|$
Distance to $-3q$ : $|x-d|$
Write Coulomb’s law with these distances:

F_{\text{from }q} = \frac{k \cdot (2q)(q)}{x^2}

F_{\text{from }-3q} = \frac{k \cdot (2q)(-3q)}{(x-d)^2}

Mistake 4: Ignoring the Sign When Equating Forces

The error: Students set magnitudes equal but forget that the direction must also be opposite. This leads to wrong algebraic signs.

How to avoid:

Write the net force equation with proper signs:

\frac{k \cdot 2q \cdot q}{x^2} + \frac{k \cdot 2q \cdot (-3q)}{(x-d)^2} = 0

Simplify:

\frac{1}{x^2} - \frac{3}{(x-d)^2} = 0

Solve:

\frac{1}{x^2} = \frac{3}{(x-d)^2}

|x-d| = \sqrt{3}\,|x|

Mistake 5: Forgetting to Check Which Solution Is Valid

The error: Students solve the equation and get two mathematical solutions, but don't check whether each lies in a physically possible region.

How to avoid:

Solve $|x-d| = \sqrt{3}\,|x|$ $∣ x - d ∣ = 3 ∣ x ∣$ :
- Case 1: $x-d = \sqrt{3}\,x$ → $x = \frac{d}{1-\sqrt{3}}$ (negative, left of $q$ — valid)
- Case 2: $x-d = -\sqrt{3}\,x$ → $x = \frac{d}{1+\sqrt{3}}$ (positive, between $q$ and $-3q$ — invalid because forces would be in the same direction there)
Always test the direction of forces for each candidate position.

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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 22 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$P=+25 \mu C$
›Reveal solutionSolution
For two opposite charges, the net electric field is zero outside the smaller charge on the line joining them. Solving $E_P = E_Q$ gives distances 20 cm from Q and 25 cm from P, so option (C) is correct.

Concept & Intuition

The electric field due to a point charge is $E = k|q|/r^2$ , directed away from a positive charge and toward a negative charge. Here we have $+25\,\mu\text{C}$ and $-16\,\mu\text{C}$ . Because the charges are opposite, the fields point in opposite directions between the charges, but they can never cancel there because the magnitudes are different and the directions are opposite — the net field is nonzero.

The only place where the fields can cancel is outside the smaller charge (the $-16\,\mu\text{C}$ ), because there the fields from both charges point in opposite directions (away from the positive, toward the negative) and the distances can be adjusted to make magnitudes equal.

We set up the condition $|E_P| = |E_Q|$ and solve for the distances.

Step-by-step solution

Define the geometry

Let the charges be on the x‑axis: $P$ at $x=0$ with $+25\,\mu\text{C}$ , $Q$ at $x=5$ cm with $-16\,\mu\text{C}$ .

Let the zero‑field point be at distance $x$ from $P$ (so $x$ cm from $P$ ).

If the point is to the right of $Q$ , then distance from $P$ is $x$ and from $Q$ is $x-5$ (both in cm).

Write the field magnitudes

$E_P = \frac{k \cdot 25}{(x)^2}, \qquad E_Q = \frac{k \cdot 16}{(x-5)^2}$
(We drop the sign because we only care about magnitude; direction is handled by the geometry.)

Set magnitudes equal for cancellation

$\frac{25}{x^2} = \frac{16}{(x-5)^2}$
Take square roots (positive distances):
$\frac{5}{x} = \frac{4}{x-5}$

Solve for $x$ Cross‑multiply:

$5(x-5) = 4x \quad\Rightarrow\quad 5x - 25 = 4x \quad\Rightarrow\quad x = 25\ \text{cm}$
So the point is 25 cm from $P$ . …
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COMEDK 2026Set 2026-M1 markMCQ
$45^{\circ}$
›Reveal solutionSolution
The key idea is that the angle remains the same because the electric force is reduced by the dielectric constant exactly as much as the effective weight is reduced by buoyancy. Solving the equilibrium condition gives the dielectric constant as 1.5.

We have two identical charged spheres hanging from strings of equal length, making an angle of $45^\circ$ with each other. That means each string makes an angle of $22.5^\circ$ with the vertical. When immersed in a liquid, the angle stays the same. This tells us that the ratio of the electric repulsion to the effective weight (weight minus buoyancy) must be the same in air and in the liquid.

Concept and intuition:

In air, the spheres repel each other with a force $F_e = \frac{k q^2}{r^2}$ (where $k = 1/(4\pi\varepsilon_0)$ ). In the liquid, the electric force is reduced by the dielectric constant $\kappa$ : $F_e' = \frac{k q^2}{\kappa r^2}$ . Meanwhile, the weight of each sphere is $mg$ , but in the liquid there is an upward buoyant force $F_b = \rho_{\text{liquid}} V g$ . The effective weight becomes $mg - F_b = (\rho_{\text{sphere}} - \rho_{\text{liquid}}) V g$ . Since the angle doesn’t change, the ratio of electric force to effective weight must be the same in both cases. That gives us an equation to solve for $\kappa$ .

Let’s work through it step by step.

Set up the equilibrium in air. Each sphere has mass $m = \rho_s V$ , where $\rho_s = 2.4 \, \text{g/cm}^3$ and $V$ is its volume. The tension in the string has vertical component $T \cos\theta = mg$ and horizontal component $T \sin\theta = F_e$ , where $\theta = 22.5^\circ$ (half the angle between the strings). Dividing:

$\tan\theta = \frac{F_e}{mg}.$

Set up the equilibrium in the liquid. The effective weight is $mg - F_b = (\rho_s - \rho_l) V g$ , with $\rho_l = 0.8 \, \text{g/cm}^3$ (relative density 0.8 means density 0.8 g/cm³). The electric force is now $F_e' = \frac{F_e}{\kappa}$ because the medium reduces the Coulomb force by the dielectric constant. The angle is the same, so:

$\tan\theta = \frac{F_e'}{(\rho_s - \rho_l) V g}.$

Equate the two expressions for $\tan\theta$ . From air: $\tan\theta = \frac{F_e}{\rho_s V g}$ . From liquid: $\tan\theta = \frac{F_e / \kappa}{(\rho_s - \rho_l) V g}$ . …
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COMEDK 2026Set 2026-M1 markMCQ
$\sqrt{2} C, \sqrt{2} C$
›Reveal solutionSolution

The two equal $\sqrt{2}\,\text{C}$ charges sit at the ends of the hypotenuse, collinear through the midpoint $P$ ; their fields cancel there. Only the $-2\,\text{C}$ charge remains, giving $7.2\times10^{10}\,\text{N C}^{-1}$ .

Geometry

$P$ is equidistant from all three vertices of a right-angled triangle, so it is the circumcentre — the midpoint of the hypotenuse — with $XP = YP = ZP = 0.5\,\text{m}$ .

Fields of the two positive charges cancel

The two $+\sqrt{2}\,\text{C}$ charges lie at opposite ends of the hypotenuse, so $P$ is exactly between them and on the line joining them. Each produces a field of equal magnitude directed away from itself; being back-to-back along the same line, they are equal and opposite at $P$ and cancel:

$E_{\sqrt{2}} = \frac{kq}{r^2} = \frac{9\times10^{9}\times\sqrt{2}}{(0.5)^2}\;\text{each, but they sum to zero.}$ …
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KCET 2025Set D-41 markMCQ
$Q. Which of the following statements is not true? (A) Work done to move a charge on an equipotential surface is not zero (B) Equipotential surfaces are the surfaces where the potential is constant (C) Equipotential surfaces for a uniform electric field are parallel and equidistant from each other (D) Electric field is always perpendicular to an equipotential surfaces.$
›Reveal solutionSolution

Work on an equipotential surface is $W = q\,\Delta V = 0$ by definition, so the statement that it is 'not zero' is the false one.

Step 1 — Definition of an equipotential surface.

An equipotential surface is a surface on which the electrostatic potential $V$ has the same value at every point. This makes statement (B) true — it is simply the definition.

Step 2 — Test statement (A): work done moving a charge on such a surface.

The work done by an external agent in moving a charge $q$ from point $A$ to point $B$ is
$W = q\,(V_B - V_A) = q\,\Delta V$
If $A$ and $B$ both lie on the same equipotential surface, then $V_A = V_B$ , so
$\Delta V = 0 \quad\Longrightarrow\quad W = q \times 0 = \boxed{0}$
The work is zero. Statement (A) asserts it is not zero — therefore (A) is FALSE, and since the question asks which statement is not true, (A) is the answer.

Step 3 — Confirm (D) is true (field ⊥ surface).

If $\vec{E}$ had any component $E_\parallel$ along the surface, moving a charge a distance $d\ell$ along the surface would do work $W = qE_\parallel\,d\ell \neq 0$ , which would mean a potential difference between two points of the surface — contradicting its being equipotential. Hence $E_\parallel = 0$ , i.e. $\vec{E}$ is always normal (perpendicular) to the equipotential surface. (D) is true.

Step 4 — Confirm (C) is true (uniform field case). …
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COMEDK 2025Set 2025-M1 markMCQ
$q_1$
›Reveal solutionSolution
When two identical charged conductors touch, they share charge equally, and the electrostatic force after separation is always greater than or equal to the original force unless one charge was zero. The correct option is (D).

The key idea here is charge redistribution upon contact. When two identical conducting spheres touch, they become a single conductor, and charge flows until both have the same potential. Since they are identical, this means they end up with equal charges. The force between them after separation depends on the product of the new charges compared to the original product.

Original force The initial electrostatic force between the balls is given by Coulomb’s law:

$F_{\text{initial}} = k \frac{q_1 q_2}{r^2}$
where $k = \frac{1}{4\pi\varepsilon_0}$ .

Charge redistribution on contact When the two identical conducting balls touch, the total charge $q_1 + q_2$ is shared equally because they are identical in size and shape. So each ball ends up with:

$q' = \frac{q_1 + q_2}{2}$

Force after separation After they are separated back to the same distance $r$ , the new force is:

$F_{\text{new}} = k \frac{(q')^2}{r^2} = k \frac{\left(\frac{q_1+q_2}{2}\right)^2}{r^2} = k \frac{(q_1+q_2)^2}{4r^2}$

Compare $F_{\text{new}}$ and $F_{\text{initial}}$ We compare the products:

$\text{Initial product: } P_i = q_1 q_2$ $\text{New product: } P_n = \frac{(q_1+q_2)^2}{4}$
The difference is:
$P_n - P_i = \frac{(q_1+q_2)^2}{4} - q_1 q_2 = \frac{q_1^2 + 2q_1q_2 + q_2^2 - 4q_1q_2}{4} = \frac{q_1^2 - 2q_1q_2 + q_2^2}{4} = \frac{(q_1 - q_2)^2}{4}$
Since $(q_1 - q_2)^2 \ge 0$ , we have $P_n \ge P_i$ . Equality only if $q_1 = q_2$ .

Conclusion for the given problem …
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KCET 2024Set D-21 markMCQ
$2aq$
›Reveal solutionSolution

Gauss's law counts only the net enclosed charge, and a dipole's net charge is zero.

Step 1 — State Gauss's law

For any closed surface $S$ ,
$\oint_S \vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0}$
The crucial point of this law is that the right-hand side depends only on the algebraic sum of the charges inside — not on their positions, not on the size or shape of the surface, and not on any charges outside.

Step 2 — Find the charge enclosed by the sphere

A dipole of moment $p = 2aq$ consists of two point charges $+q$ and $-q$ separated by $2a$ . Since the whole dipole lies inside the spherical surface, both charges are enclosed:
$q_{\text{enc}} = (+q) + (-q) = 0$
Step 3 — Apply Gauss's law
$\phi = \frac{q_{\text{enc}}}{\epsilon_0} = \frac{0}{\epsilon_0} = 0$
Step 4 — Why this is not a trick …
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KCET 2024Set D-21 markMCQ
$\dfrac{\sigma}{2\epsilon_0}$
›Reveal solutionSolution
Under electrostatic conditions, a conductor’s excess charge sits only on its surface, making option (C) the correct statement.

The key idea is that in electrostatics, charges inside a conductor are free to move until they reach equilibrium. Once equilibrium is reached, the electric field inside the conductor must be zero everywhere. If there were any field, charges would keep moving. This single fact — zero internal field — forces all the other properties: the surface is an equipotential, the field just outside is perpendicular to it, and any extra charge cannot stay inside.

Let’s examine each statement carefully.

Option (A): The electric field on the surface of a charged conductor is $\dfrac{\sigma}{2\epsilon_0}$ .

This formula is actually the field due to an infinite sheet of charge. For a conductor, the field just outside the surface is $\dfrac{\sigma}{\epsilon_0}$ , not half that. Why? Because the field inside the conductor is zero, so all the flux through a Gaussian pillbox that straddles the surface comes from the outside face. That gives $\dfrac{\sigma}{\epsilon_0}$ . The $\dfrac{\sigma}{2\epsilon_0}$ result applies to a thin sheet of charge in free space, where field exists on both sides. Here, the conductor’s interior has no field, so the outside field is double that. So (A) is false.

Option (B): The electric potential inside a charged conductor is always zero.

Potential is defined relative to a reference (usually infinity). Inside a conductor in equilibrium, the field is zero, so the potential is constant throughout — but that constant need not be zero. It could be positive or negative depending on the charge and the reference. For example, a positively charged isolated conductor has a positive potential. So (B) is false.

Option (C): Any excess charge resides on the surface of the conductor. …
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KCET 2024Set D-21 markMCQ
$0.2 \times 10^{-6}\,\text{C} - \text{m}$
›Reveal solutionSolution

Get the polarisation (total dipole moment per unit volume), then divide by the applied field, since the stated unit $\text{C}^2\text{N}^{-1}\text{m}^{-2}$ means $\chi$ is defined here by $P=\chi E$ .

Step 1 — The concept: polarisation

When a dielectric is placed in an external field, each molecule acquires an induced dipole moment. The macroscopic measure of this is the polarisation vector
$P = \frac{\text{net dipole moment}}{\text{volume}}$
and for a linear dielectric $P$ is proportional to the field, the constant of proportionality being the electric susceptibility.

Step 2 — Volume of the cube
$a = 1\ \text{cm} = 10^{-2}\ \text{m} \quad\Rightarrow\quad V = a^3 = \left(10^{-2}\right)^3 = 10^{-6}\ \text{m}^3$
Step 3 — Total dipole moment

The $100$ molecules each carry $p = 0.2\times10^{-6}\ \text{C\,m}$ and (being induced by the same field) are aligned, so their moments simply add:
$p_{\text{total}} = N p = 100 \times 0.2\times10^{-6} = 2\times10^{-5}\ \text{C\,m}$
Step 4 — Polarisation
$P = \frac{p_{\text{total}}}{V} = \frac{2\times10^{-5}}{10^{-6}} = 20\ \text{C\,m}^{-2}$
Step 5 — Susceptibility …
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COMEDK 2024Set 2024-A1 markMCQ
$-\mathrm{q}$
›Reveal solutionSolution

Two of the three equal charges sit at opposite ends of a diameter and their fields at O cancel; only the third charge (at B) is left, so the net field points towards OB - option (C).

The field at O is the vector sum of the fields of the three charges. Each charge is $-q$ at a distance equal to the radius $a$ (diameter $=2a$ ), and for a negative charge the field at O points toward that charge.

Charges A and C lie at opposite ends of a diameter, so their fields at O are equal in magnitude and exactly opposite in direction:
$\vec{E}_A+\vec{E}_C=\frac{Kq}{a^2}\,\hat{u}_{OA}+\frac{Kq}{a^2}\,\hat{u}_{OC}=\vec{0}.$
Charge B is then unopposed, so the resultant is its field alone, directed from O toward B:

$\vec{E}_O=\frac{Kq}{a^2}\ \text{along } OB.$ …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{M}$
›Reveal solutionSolution

Transferring $30\%$ of $N$ 's charge leaves both magnitudes at $0.7q$ , so the force scales by $0.7\times0.7=0.49$ , giving $\dfrac{49}{100}F$ .

Initial force between $+q$ and $-q$ separated by $r$ :
$F=\frac{1}{4\pi\varepsilon_0}\frac{q\cdot q}{r^{2}}.$
$30\%$ of $N$ 's charge is $0.3q$ , transferred from $N$ to $M$ :

$q_M=q-0.3q=0.7q,\qquad q_N=q-0.3q=0.7q\ \text{(magnitude).}$ …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{A}, \mathrm{B}$
›Reveal solutionSolution

The two equal positive charges give $10000$ N/C net along the axis and the negative charge $5000$ N/C the other way; resultant $=5000$ N/C.

With $r=6.0$ cm $=0.06$ m, $k=9\times10^9$ :
$E_{+4}=\frac{(9\times10^9)(4\times10^{-9})}{(0.06)^2}=\frac{36}{0.0036}=10000\text{ N/C}$
$E_{-2}=\frac{(9\times10^9)(2\times10^{-9})}{(0.06)^2}=5000\text{ N/C}$ …
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COMEDK 2024Set 2024-M1 markMCQ
$q$
›Reveal solutionSolution
The key idea is that the electric field at the centre of a regular pentagon due to five equal charges is zero by symmetry. Removing one charge leaves a net field equal in magnitude to the field of that missing charge alone. Replacing it with an equal opposite charge doubles that field. The ratio of magnitudes is therefore 1 : 2, corresponding to option (D).

Concept and intuition:

For a regular pentagon, the centre is equidistant from all vertices. If five equal charges are placed at the vertices, the vector sum of the electric fields at the centre is zero — each field is cancelled by the others due to symmetry. This is the same principle that makes the field at the centre of a regular polygon with equal charges zero.

Now, if we remove one charge, the symmetry is broken. The net field at O is exactly the field that the missing charge would have produced (since the remaining four still sum to the negative of that one field). If we instead replace the removed charge with an equal opposite charge, that charge now contributes a field in the same direction as the field from the missing charge (because the sign flips), so the total field becomes twice the field of a single charge. The ratio of the magnitudes (first case : second case) is therefore 1 : 2.

Step-by-step reasoning:

Original configuration (five equal charges): The centre O is equidistant from all vertices. Let this distance be $r$ (the circumradius). Each charge $q$ produces a field of magnitude $E_0 = \frac{kq}{r^2}$ at O, directed radially outward (if $q>0$ ). Because the pentagon is regular, the five vectors are equally spaced in direction by $72^\circ$ . Their vector sum is zero:

$\vec{E}_{\text{total}} = \vec{0}.$

First scenario: charge at A is removed. Now only four charges remain (at B, C, D, E). Their net field at O is no longer zero. However, since the original five fields summed to zero, the sum of the four remaining fields must be exactly the negative of the field that was produced by the charge at A:

$\vec{E}_{\text{four}} = -\vec{E}_A.$
The magnitude is therefore $|\vec{E}_{\text{four}}| = E_0$ .

Second scenario: charge at A is replaced with an equal opposite charge $-q$ . …
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Q.Two charges qqq and −3q-3q−3q are placed fixed on xxx-axis separated by distance ddd. Where should a third charge 2q2q2q be placed such that it will not experience any force?

1. Analyze the possible regions for zero net force

2. Set up the force equations for Region I

3. Equate the magnitudes of the forces …

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Q.Two charges $q$ and $-3q$ are placed fixed on $x$ -axis separated by distance $d$ . Where should a third charge $2q$ be placed such that it will not experience any force?

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