Derive an expression for the torque acting on an electric dipole of dipole moment $\vec{p}$ placed in a uniform electric field $\vec{E}$. Write the direction along which the torque acts. OR Derive an expression for the electric field at a point on the axis of an electric dipole of dipole moment $\vec{p}$. Also write its expression when the distance $r \gg$ the length 'a' of the dipole.

Derive an expression for the torque acting on an electric dipole of dipole moment vecp placed in a uniform ele

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\vec{p}

(OR)

\vec{p}

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Torque On Dipole

Torque on a Dipole — From Intuition to the Formula

Imagine a bar magnet placed in a uniform magnetic field. You know that the north pole gets pulled one way and the south pole the opposite way. If the magnet is not aligned with the field, these two equal and opposite forces create a twist — a torque — that tries to rotate the magnet until it lines up with the field. That's the core idea.

The same thing happens with an electric dipole (two equal and opposite charges $+q$ and $-q$ separated by a small distance $d$ ) placed in a uniform electric field $\vec{E}$ . The two charges experience forces in opposite directions, and unless the dipole is already parallel to the field, those forces produce a torque.

Step 1: The Forces on the Two Charges

Let the dipole moment $\vec{p}$ point from the negative charge to the positive charge, with magnitude $p = qd$ .

In a uniform electric field $\vec{E}$ :

The positive charge $+q$ feels a force $\vec{F}_+ = +q\vec{E}$ (in the direction of $\vec{E}$ ).
The negative charge $-q$ feels a force $\vec{F}_- = -q\vec{E}$ (opposite to $\vec{E}$ ).

These two forces are equal in magnitude but opposite in direction. They form a couple — a pair of equal, opposite, parallel forces that do not share the same line of action. A couple always produces a pure torque, with no net force.

Step 2: Why a Torque Appears

If the dipole is at an angle $\theta$ to the field, the two forces are not along the same line. They are separated by the perpendicular distance between their lines of action. That perpendicular distance is $d \sin\theta$ , where $d$ is the separation between the charges.

The torque $\tau$ due to a couple is:

\tau = (\text{force magnitude}) \times (\text{perpendicular distance between forces})

Here:

Force magnitude on each charge: $F = qE$
Perpendicular distance: $d \sin\theta$

So:

\tau = (qE) \times (d \sin\theta) = qd \, E \sin\theta

But $qd = p$ , the magnitude of the dipole moment. Therefore:

\tau = pE \sin\theta

Step 3: The Vector Form

Torque is a vector — it has a direction. The direction of the torque is perpendicular to both $\vec{p}$ and $\vec{E}$ , following the right-hand rule. The complete vector equation is:

\vec{\tau} = \vec{p} \times \vec{E}

The magnitude is $|\vec{\tau}| = pE \sin\theta$ , where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$ .

Step 4: What the Torque Does

When $\theta = 0^\circ$ (dipole aligned with the field): $\sin 0 = 0$ , so $\tau = 0$ . The dipole is in stable equilibrium — if you nudge it slightly, the torque brings it back.
When $\theta = 90^\circ$ (dipole perpendicular to the field): $\sin 90^\circ = 1$ , so torque is maximum: $\tau_{\text{max}} = pE$ . …

Why this formula?

Torque on a Dipole in a Uniform Electric Field

Let's build this from first principles — understanding why the torque formula is what it is, not just memorizing it.

What is a Dipole?

A dipole consists of two equal and opposite charges $+q$ and $-q$ , separated by a small distance $2a$ (or $d$ ). The dipole moment vector is:

\vec{p} = q \cdot \vec{d}

where $\vec{d}$ points from $-q$ to $+q$ , and $|\vec{d}| = 2a$ .

The Physical Situation

Place this dipole in a uniform external electric field $\vec{E}$ . Uniform means the field has the same magnitude and direction everywhere.

The $+q$ charge experiences a force: $\vec{F}_+ = +q\vec{E}$
The $-q$ charge experiences a force: $\vec{F}_- = -q\vec{E}$

These two forces are equal in magnitude but opposite in direction.

Why is there a Torque?

Since the forces are equal and opposite, the net force on the dipole is zero:

\vec{F}_{\text{net}} = q\vec{E} + (-q\vec{E}) = 0

So the dipole won't accelerate linearly. But — crucially — the two forces act at different points in space (the two charges are separated). This creates a couple (a pair of equal, opposite, parallel forces not acting along the same line). A couple always produces a torque (rotational effect).

Deriving the Torque Magnitude

Let the dipole be oriented at an angle $\theta$ with respect to the field $\vec{E}$ .

The line joining the charges makes angle $\theta$ with $\vec{E}$ .
The perpendicular distance between the lines of action of the two forces is the "lever arm."

Step 1: The force on each charge is $qE$ .

Step 2: The perpendicular distance between the two forces is:

\text{Lever arm} = 2a \sin\theta

Why $\sin\theta$ ? Because the separation vector $\vec{d}$ is at angle $\theta$ to $\vec{E}$ . The component of $\vec{d}$ perpendicular to $\vec{E}$ is $d\sin\theta = 2a\sin\theta$ .

Step 3: Torque = Force × Perpendicular distance (for one force about the midpoint):

\tau = (qE) \times (2a \sin\theta)

Step 4: But $q \times 2a = p$ , the dipole moment magnitude. So:

\boxed{\tau = pE \sin\theta}

Vector Form — The Full Picture

Torque is a vector. Its direction is given by the right-hand rule: it tends to rotate the dipole toward alignment with the field.

The vector form captures both magnitude and direction:

\boxed{\vec{\tau} = \vec{p} \times \vec{E}}

Magnitude: $|\vec{\tau}| = pE \sin\theta$ (as derived)
Direction: Perpendicular to both $\vec{p}$ and $\vec{E}$ , given by the cross product rule.

--- …

Part (b)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.

--- …

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A dipole in a uniform field feels a couple: $\vec\tau=\vec p\times\vec E$ , magnitude $pE\sin\theta$ , perpendicular to both.
On the axis $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2pr}{(r^2-a^2)^2}$ along $\vec p$ , which for $r\gg a$ becomes $\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ .

Part (a) — Torque on a dipole in a uniform field

An electric dipole is two equal, opposite charges $+q$ and $-q$ a distance $2a$ apart; its moment is $\vec p=q(2a)\hat n$ , pointing from $-q$ to $+q$ .

In a uniform field $\vec E$ making angle $\theta$ with $\vec p$ :

Force on $+q$ : $\vec F_+=+q\vec E$
Force on $-q$ : $\vec F_-=-q\vec E$

The forces are equal and opposite, so the net force is zero — the dipole does not translate. But they act along different lines, forming a couple that rotates the dipole.

\tau=\text{force}\times\text{arm}=(qE)\times(2a\sin\theta)=(q\cdot 2a)E\sin\theta=pE\sin\theta.

In vector form,

$\boxed{\;\vec\tau=\vec p\times\vec E\;}$ …

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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 45 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 A1 markMCQ
$Q. When an electric dipole p is placed in a uniform electric field E, then at what angle between p and E, the value of torque will be maximum? (A) 0° (B) 45° (C) 90° (D) 180°$
›Reveal solutionSolution

The torque on a dipole is τ = pE sinθ, which peaks when the dipole is perpendicular to the field (θ = 90°).

The torque on an electric dipole of moment $p$ in a uniform field $E$ is

$\tau = pE\sin\theta$ …

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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 55/4/11 markMCQ
$30^\circ$
›Reveal solutionSolution

The torque on a magnetic dipole is $\tau = mB\sin\theta$ , maximum when perpendicular ( $\theta = 90^\circ$ ). When torque drops to half its initial value, $\sin\theta = \frac{1}{2}$ , giving $\theta = 30^\circ$ — so the magnet rotates through $60^\circ$ .

Understanding torque on a magnetic dipole

When a bar magnet (magnetic dipole of moment $m$ ) sits in a uniform magnetic field $B$ , it experiences a torque that tries to align it with the field. The magnitude of this torque depends on how misaligned the dipole is:
$\tau = mB\sin\theta$
where $\theta$ is the angle between the magnetic moment vector and the field direction.

The torque is maximum when the dipole is perpendicular to the field ( $\theta = 90^\circ$ , so $\sin\theta = 1$ ), and zero when aligned ( $\theta = 0^\circ$ ). This makes physical sense: the restoring couple is strongest when the magnet is sideways to the field lines.
$\tau = mB\sin\theta$
Step-by-step solution

1. Identify the initial configuration

The magnet starts "at right angles to a uniform magnetic field," meaning the magnetic moment makes an angle $\theta_i = 90^\circ$ with the field. The initial torque is:
$\tau_i = mB\sin 90^\circ = mB$
2. Set up the condition after rotation

After rotating the magnet, the torque becomes half the initial value:
$\tau_f = \frac{\tau_i}{2} = \frac{mB}{2}$
Let the new angle between the magnetic moment and field be $\theta_f$ . Then:
$mB\sin\theta_f = \frac{mB}{2}$
3. Solve for the final angle

Dividing both sides by $mB$ :
$\sin\theta_f = \frac{1}{2}$
This gives: …
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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
$2\times10^5$
(a) 5 mC
(b) 4 mC
(c) 7 mC
(d) 8 mC
›Reveal solutionSolution

Computing the dipole moment from $\tau=pE\sin\theta$ and then the charge from $p=qd$ gives $q=8$ mC.

Working

Torque on a dipole in a uniform field: $\tau = pE\sin\theta$ .

Given $\tau=8$ Nm, $E=2\times10^5\ \text{NC}^{-1}$ , $\theta=30°$ :

$p = \dfrac{\tau}{E\sin\theta} = \dfrac{8}{(2\times10^5)(0.5)} = \dfrac{8}{1\times10^5} = 8\times10^{-5}\ \text{Cm}$

…

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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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Q.Derive an expression for the torque acting on an electric dipole of dipole moment $\vec{p}$ placed in a uniform electric field $\vec{E}$ . Write the direction along which the torque acts.

Part (a)Concept understanding — Torque On Dipole

Torque on a Dipole — From Intuition to the Formula

Step 1: The Forces on the Two Charges

Step 2: Why a Torque Appears

Step 3: The Vector Form

Step 4: What the Torque Does

Why this formula?

Torque on a Dipole in a Uniform Electric Field

What is a Dipole?

The Physical Situation

Why is there a Torque?

Deriving the Torque Magnitude

Vector Form — The Full Picture

Part (b)Concept understanding — Coulomb Force Superposition

Coulomb Force Superposition – From Intuition to Precision

The Precise Statement

Why This Matters (and a Common Trap)

The Deeper Reason

Why this formula?

Coulomb Force Superposition — Why the Formula Holds

The Key Formula

Why This Works — The Physical Reasoning

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

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

3. The Electric Field is Linear

The Crucial Assumption (Why It's Not Trivial)

Part (a) — Torque on an electric dipole

Part (a) — Torque on a dipole in a uniform field

Understanding torque on a magnetic dipole

Step-by-step solution

Q.Derive an expression for the torque acting on an electric dipole of dipole moment p⃗\vec{p}p​ placed in a uniform electric field E⃗\vec{E}E. Write the direction along which the torque acts.

Part (a) — Torque on a dipole in a uniform field

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Q.Derive an expression for the torque acting on an electric dipole of dipole moment $\vec{p}$ placed in a uniform electric field $\vec{E}$ . Write the direction along which the torque acts.

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