An electric dipole is placed at an alignment angle of 30° with an electric field of $2\times10^5$ NC$^{-1}$. It experiences a torque equal to 8 Nm. The charge on the dipole if the dipole length is 1 cm is : (a) 5 mC (b) 4 mC (c) 7 mC (d) 8 mC

An electric dipole is placed at an alignment angle of 30° with an electric field of 2×10⁵ NC⁻¹. It experiences

Question 113 of 122

2\times10^5

(a) 5 mC

(b) 4 mC

(d) 8 mC

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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.

--- …

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

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 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 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 2023Set ANNUAL1 markMCQ
$Q. Torque acting on electric dipole of dipole moment p⃗ placed in uniform electric field E⃗ is -$
(a) p⃗ × E⃗
(b) p⃗ . E⃗
(c) p⃗ × (E⃗ × p⃗)
(d) E . (p⃗/p²)
›Reveal solutionSolution
An electric dipole in a uniform field feels equal and opposite forces on its two charges, forming a couple whose torque is the vector cross product of the dipole moment and the field.

Why: For a dipole of charges $+q$ and $-q$ separated by $\vec d$ (dipole moment $\vec p=q\vec d$ ) placed in a uniform field $\vec E$ , the force on $+q$ is $q\vec E$ and on $-q$ is $-q\vec E$ — equal, opposite, and (since the field is uniform) acting at different points, so they form a couple with no net force but a net torque.

Steps:

Torque of a couple = force × perpendicular distance between the lines of action. …
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CBSE 2017Set ANNUAL1 markMCQ
$Q. The torque acting on an electric dipole of dipole moment P placed at an angle 90° to the electric field E will be: (A) PE (B) PE cos θ (C) PE / sin θ (D) Zero$
›Reveal solutionSolution

Dipole torque is $\tau = PE\sin\theta$ ; at $90^\circ$ this reduces to $PE$ .

An electric dipole of moment $\vec P$ placed in a uniform field $\vec E$ experiences a torque

$\vec\tau = \vec P \times \vec E, \quad |\tau| = PE\sin\theta$

where $\theta$ is the angle between $\vec P$ and $\vec E$ .

Substituting $\theta = 90^\circ$ : $\sin 90^\circ = 1$ , so

$\tau = PE \times 1 = PE$ …

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Q.An electric dipole is placed at an alignment angle of 30° with an electric field of 2×1052\times10^52×105 NC−1^{-1}−1. It experiences a torque equal to 8 Nm. The charge on the dipole if the dipole length is 1 cm is :

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Q.An electric dipole is placed at an alignment angle of 30° with an electric field of $2\times10^5$ NC $^{-1}$ . It experiences a torque equal to 8 Nm. The charge on the dipole if the dipole length is 1 cm is :

Unlock everything free for 14 days