Consider an electric dipole comprising two charges +q and -q each with mass m, separated by a fixed

Q: Why is the net force zero if the dipole is rotating?

The electric field is uniform, meaning it exerts equal and opposite forces on the $+q$ and $-q$ charges. These forces cancel each other out translationally, so the center of mass doesn't accelerate. However, because their lines of action are different, they create a couple (torque) that causes rotation.

Q: Why is the angle between $\vec{p}$ and $\vec{E}$ taken as $(90^\circ - \theta)$?

The dipole moment $\vec{p}$ makes an angle $\theta$ with the x-axis ($\hat{i}$). The electric field $\vec{E}$ is along the y-axis ($\hat{j}$), which is at $90^\circ$ to the x-axis. Therefore, the angle between $\vec{p}$ and $\vec{E}$ is $90^\circ - \theta$.

Q: Can we use potential energy to find the change in kinetic energy?

Absolutely! The work done by the electric field is equal to the negative change in potential energy: $W = -\Delta U = U_i - U_f$. Since $U = -\vec{p} \cdot \vec{E} = -pE \cos(90^\circ - \theta) = -pE \sin\theta$, you get $W = 0 - (-pE \sin\theta_f) = pE \sin\theta_f$.

Q: Why does the dipole continue to rotate after $t_f$?

According to Newton's First Law for rotation, an object will maintain its state of rotational motion unless acted upon by a net external torque. Once the field is off, torque is zero, so it spins with constant angular velocity.

Q: What is the moment of inertia of the dipole?

The dipole consists of two point masses $m$ separated by a distance $d$. They rotate about their center of mass, which is exactly in the middle. So each mass is at a distance $d/2$ from the axis. $I = m(d/2)^2 + m(d/2)^2 = md^2/2$.

Question

Consider an electric dipole comprising two charges $+q$ and $-q$ each with mass $m$, separated by a fixed distance $d$ and initially at rest with its dipole moment pointing along $\hat{i}$. A uniform electric field $E\hat{j}$ is turned on at time $t = 0$ and it is turned off at $t = t_f$, when the dipole moment makes an angle $	heta_f$ with $\hat{i}$. Neglecting any sources of energy loss, correct option(s) is/are:

Options:

(A) The center of mass of the dipole is deflected towards $\hat{j}$ in the presence of the field.

(B) If the magnitude of the final angular velocity $\omega_f = \sqrt{\frac{2qE}{md}}$ then $	heta_f = \frac{\pi}{6}$

(C) If $	heta_f = \pi/4$, then the change in kinetic energy of the dipole is given by $2\sqrt{3} qEd$.

(D) For $	heta_f = \pi/4$, the dipole rotates around its center of mass with a constant angular velocity after $t > t_f$.

Sigma Percentile · Accepted Answer

### The Calm Before the Storm

Imagine a perfectly balanced electric dipole resting peacefully along the x-axis. It consists of two charges, $+q$ and $-q$, separated by a distance $d$. Suddenly, at $t=0$, a uniform electric field $\vec{E} = E\hat{j}$ is switched on, pointing straight up along the y-axis. This sudden jolt of energy is going to disrupt our dipole's peaceful existence. Let's break down exactly how it responds.

### Translational Equilibrium: Why the Center Holds

The very first thing we must check is whether the dipole as a whole will start drifting away. To do this, we look at the **net force**.

The positive charge $+q$ experiences a force pushing it upwards:

$$ \vec{F}_+ = qE\hat{j} $$

Meanwhile, the negative charge $-q$ experiences an equal and opposite force pulling it downwards:

$$ \vec{F}_- = -qE\hat{j} $$

When we sum these up, the magic happens:

$$ \vec{F}_{net} = qE\hat{j} - qE\hat{j} = 0 $$

Because the net force is zero, the center of mass of the dipole will not accelerate in any direction. It stays perfectly anchored. This immediately tells us that **Option A is incorrect**.

### The Twist: Calculating the Torque

Even though the center of mass isn't moving, the forces are acting at different points. This creates a couple, or a **torque**, which will cause the dipole to spin.

Let's calculate this torque when the dipole is at an arbitrary angle $	heta$ with the x-axis. The dipole moment vector $\vec{p}$ points from $-q$ to $+q$:

$$ \vec{p} = qd(\cos	heta \hat{i} + \sin	heta \hat{j}) $$

The torque $\vec{	au}$ is the cross product of the dipole moment and the electric field:

$$ \vec{	au} = \vec{p} 	imes \vec{E} = (qd\cos	heta \hat{i} + qd\sin	heta \hat{j}) 	imes (E\hat{j}) $$

Since $\hat{i} 	imes \hat{j} = \hat{k}$ and $\hat{j} 	imes \hat{j} = 0$, we get:

$$ \vec{	au} = qEd\cos	heta \hat{k} $$

This torque is what drives the rotational motion of our dipole.

### Rotational Inertia

Before we can find out how fast it spins, we need to know its resistance to spinning—its **moment of inertia** ($I$). The dipole rotates about its center of mass, which is exactly halfway between the two charges. Each mass $m$ is at a distance of $d/2$ from the rotation axis.

$$ I = m\left(\frac{d}{2}ight)^2 + m\left(\frac{d}{2}ight)^2 = \frac{md^2}{2} $$

### The Work-Energy Connection

As the electric field twists the dipole, it does work. According to the **Work-Energy Theorem**, this work translates directly into the dipole's rotational kinetic energy.

Let's integrate the torque from the initial angle ($0$) to the final angle ($	heta_f$):

$$ W = \int_0^{	heta_f} 	au d	heta = \int_0^{	heta_f} qEd\cos	heta d	heta $$

Evaluating the integral of cosine gives us sine:

$$ W = qEd[\sin	heta]_0^{	heta_f} = qEd\sin	heta_f $$

This work equals the final kinetic energy $\frac{1}{2}I\omega_f^2$. So, our master equation is:

$$ \frac{1}{2}I\omega_f^2 = qEd\sin	heta_f $$

### Testing the Hypotheses

Now we are fully equipped to test the remaining options. Let's look at **Option B**. It proposes a specific final angular velocity:

$$ \omega_f = \sqrt{\frac{2qE}{md}} $$

Let's plug this into our master equation:

$$ \frac{1}{2}\left(\frac{md^2}{2}ight)\left(\frac{2qE}{md}ight) = qEd\sin	heta_f $$

Watch how beautifully the terms cancel out! The $m$ and one $d$ vanish, leaving us with:

$$ \frac{qEd}{2} = qEd\sin	heta_f $$

Dividing both sides by $qEd$, we find:

$$ \sin	heta_f = \frac{1}{2} \implies 	heta_f = \frac{\pi}{6} $$

This matches Option B perfectly! **Option B is correct.**

Next, let's test **Option C**. It suggests that if $	heta_f = \pi/4$, the kinetic energy is $2\sqrt{3}qEd$. Let's use our kinetic energy formula:

$$ \Delta KE = qEd\sin\left(\frac{\pi}{4}ight) = \frac{qEd}{\sqrt{2}} $$

This is nowhere near $2\sqrt{3}qEd$. Therefore, **Option C is incorrect**.

### The Aftermath: Conservation of Angular Momentum

Finally, let's consider **Option D**. At time $t_f$, the electric field is abruptly turned off. What happens to the spinning dipole?

Without the electric field, the torque instantly drops to zero ($\vec{	au} = 0$). According to Newton's First Law of Rotational Motion, if the net external torque is zero, the angular acceleration is zero, and the angular velocity remains **constant**.

The dipole will simply continue to spin forever (since we are neglecting energy loss) at whatever $\omega_f$ it had reached. Thus, **Option D is correct**.

Select Answer:

Visualized Solution

\text{Initial Setup}

\text{Translational Equilibrium}

\text{Torque on Dipole}

\text{Moment of Inertia}

\text{Work-Energy Theorem}

\text{Evaluating Option B}

\text{Evaluating Option C}

\text{Evaluating Option D}

The Sigma Insight: Electric Dipole

The Calm Before the Storm

Translational Equilibrium

The Twist

Rotational Inertia

The Work-Energy Connection

Testing the Hypotheses

The Aftermath

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