Q.A molecule of a substance has a permanent electric dipole moment of magnitude $10^{-29}\ \text{C m}$. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude $10^{6}\ \text{V m}^{-1}$. The direction of the field is suddenly changed by an angle of $60^\circ$. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.

The key idea is that torque and potential energy are two sides of the same dipole-in-field coin: torque depends on $\sin\theta$, energy on $\cos\theta$. Once you find the dipole moment from the torque, the energy follows directly.

1. Recall the definitions. For an electric dipole of moment $\vec{p}$ in a uniform field $\vec{E}$, the torque is

and the potential energy is

where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$.

2. Extract the given data.

   • Field strength: $E = 10^6 \, \text{N/C}$
   • Length of dipole: $4\,\text{cm} = 0.04\,\text{m}$ (though length alone isn’t directly used — we need $p = q \times \text{length}$, but we’ll find $p$ from torque)
   • Angle: $\theta = 60^\circ$
   • Torque: $\tau = 8\sqrt{3} \, \text{N·m}$

3. Find the dipole moment $p$ from the torque equation.

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$,

4. Now compute the potential energy. $$$$

Concept & Intuition

A water molecule is a permanent electric dipole: its centre of positive charge (the hydrogen side) and centre of negative charge (the oxygen side) are slightly separated. In a uniform external electric field $\vec{E}$, the dipole experiences a torque that tends to align it with the field. The work done by an external agent to rotate the dipole from alignment ($\theta = 0^\circ$) to anti-alignment ($\theta = 180^\circ$) equals the change in its potential energy. For a dipole of moment $p = q \cdot d$ (where $q$ is the magnitude of separated charge and $d$ is the separation we want), the potential energy is $U = -pE\cos\theta$. The work done for a $180^\circ$ rotation is therefore $W = \Delta U = 2pE$. From the given $W$ and $E$, we find $p$, then use the known elementary charge $e = 1.6 \times 10^{-19} \, \mathrm{C}$ to extract $d$.

Step-by-step solution

1. Relate work to dipole moment The potential energy of a dipole in a uniform field is

Rotating from $\theta = 0^\circ$ (aligned, lowest energy) to $\theta = 180^\circ$ (anti-aligned, highest energy) changes the energy by

This $\Delta U$ is the work done by the field; the work done by an external agent to overcome the field is the same magnitude. Hence

2. Solve for the dipole moment $p$ Given $W = 5 \times 10^{-25} \, \mathrm{J}$ and $E = 2.5 \times 10^4 \, \mathrm{N/C}$,

3. Relate dipole moment to charge separation …

The problem gives a charged object in a uniform electric field, released from rest. It gains kinetic energy as it moves. The natural tool here is the Work–Energy Theorem: the net work done on an object equals its change in kinetic energy. Since the only force doing work is the electric force (gravity is negligible for such a small mass over this short distance), the work done by the field directly becomes the kinetic energy.

The electric force is $\vec{F}_e = Q \vec{E}$. Because the field is uniform and along $\hat{i}$, the force is constant. The work done by a constant force over a displacement $\Delta x$ is $W = F \Delta x \cos\theta$, where $\theta$ is the angle between force and displacement. Here, the displacement is also along $\hat{i}$ (from $x=0$ to $x=0.5$ m), so the direction matters: if $Q$ is positive, the force is in the $+\hat{i}$ direction, same as displacement, so work is positive; if $Q$ is negative, force is opposite to displacement, work is negative.

Let’s go step by step.

1. Identify the knowns

   Mass $m = 2~\text{g} = 2 \times 10^{-3}~\text{kg}$ (not needed directly, since we use energy, not acceleration).

   Electric field $\vec{E} = 300~\hat{i}~\text{NC}^{-1}$.

   Displacement $\Delta x = 0.5~\text{m}$.

   Final kinetic energy $K_f = 0.12~\text{J}$, initial $K_i = 0$.

2. Apply Work–Energy Theorem

   $W_{\text{net}} = \Delta K = K_f - K_i = 0.12~\text{J}$.

   The only work is from the electric force: $W_e = Q E \Delta x$, but careful with sign. Since the object moves from $x=0$ to $x=0.5$ m, displacement vector is $+0.5~\hat{i}~\text{m}$. The force is $Q \times 300~\hat{i}~\text{N}$.

   Work $W = \vec{F} \cdot \Delta \vec{x} = (300 Q~\hat{i}) \cdot (0.5~\hat{i}) = 150 Q$ joules (with $Q$ in coulombs).

3. Set equal and solve

   $150 Q = 0.12$

   $Q = \frac{0.12}{150} = 0.0008~\text{C} = 800 \times 10^{-6}~\text{C} = 800~\mu\text{C}$. …

The key is to separate the two effects: translation (force) and rotation (torque). A uniform field gives equal and opposite forces on the two charges — zero net force, but a torque if the dipole is not aligned. A non-uniform field breaks the symmetry: the field strength (and possibly direction) differs at the two charge locations, so the forces no longer cancel.

Let’s walk through it.

1. What a dipole is

   An electric dipole consists of two equal and opposite charges $+q$ and $-q$, separated by a small distance $\mathbf{d}$ (vector from $-q$ to $+q$). Its dipole moment is $\mathbf{p} = q \mathbf{d}$.

2. Force on each charge in a general field

   Let the field at the position of $-q$ be $\mathbf{E}_-$ and at $+q$ be $\mathbf{E}_+$.

   Force on $-q$: $\mathbf{F}_- = -q \mathbf{E}_-$

   Force on $+q$: $\mathbf{F}_+ = +q \mathbf{E}_+$

   The net force on the dipole is

3. Why a non-uniform field gives a net force

   If the field is uniform, $\mathbf{E}_+ = \mathbf{E}_-$, so $\mathbf{F}_{\text{net}} = 0$.

   In a non-uniform field, $\mathbf{E}_+ \neq \mathbf{E}_-$, so the vector difference is non-zero. The dipole is pulled toward the region of stronger field (if aligned with the field) or pushed the other way (if anti-aligned). A net force always exists unless the field happens to be symmetric in a very special way — but “generally” means for an arbitrary non-uniform field, there is a net force.

4. Torque on the dipole

   Torque about the centre of the dipole is

where $\mathbf{r}_+ = \mathbf{d}/2$ and $\mathbf{r}_- = -\mathbf{d}/2$ relative to the centre.

This simplifies to

where $\mathbf{E}_{\text{avg}}$ is the average field over the dipole length. In a non-uniform field, $\mathbf{E}_{\text{avg}}$ is generally not parallel to $\mathbf{p}$, so the cross product is non-zero — a torque exists. …

Step 1 — The Carnot relation and why it holds.

For a reversible (Carnot) engine the net entropy change over a cycle is zero. The engine absorbs $Q_1$ isothermally at $T_1$ and rejects $Q_2$ isothermally at $T_2$, so

This is what makes a Carnot engine special: heat ratio = absolute temperature ratio.

Step 2 — Insert the data.

$Q_1 = 300$ cal (from the source), $Q_2 = 150$ cal (to the sink), $T_1 = 500$ K.

Step 3 — Solve.

(The units of heat cancel in the ratio, so working in calories is fine — no conversion to joules is needed.)

Step 4 — Cross-check with the efficiency. …