Animated Solution for Physics - Electrostatics: An electric dipole has a fixed dipole moment $\mathbf{p}$, which makes angle $\theta$ with respect to X-axis. When subjected to an electric field ${\mathbf{E}}_1=E\hat{\mathbf{i}}$, it experiences a torque ${\mathbf{T}}_1=\tau \hat{\mathbf{k}}$. When subjected to another electric field ${\mathbf{E}}_2=\sqrt{3}E\hat{\mathbf{j}}$, it experiences a torque ${\mathbf{T}}_2=-{\mathbf{T}}_1$. The angle $\theta$ is

Visualizing the Dipole

Imagine an electric dipole resting in a two-dimensional plane. We are told that its dipole moment vector, $\mathbf{p}$, makes an angle $\theta$ with the positive X-axis. To tackle this problem with absolute mathematical rigor, we must first express this dipole moment in its component form.

Using basic trigonometry, we can resolve $\mathbf{p}$ into its X and Y components:

This vector representation is our master key. It will allow us to compute torques without ever having to guess directions using the right-hand rule.

Electric Field Along X-axis

The fundamental law of electrostatics tells us that the torque $\mathbf{\tau }$ experienced by a dipole in an electric field $\mathbf{E}$ is given by the cross product:

In our first scenario, the dipole is subjected to an electric field ${\mathbf{E}}_1$ that points purely along the X-axis. So, ${\mathbf{E}}_1=E\hat{\mathbf{i}}$. Let's compute the resulting torque, ${\mathbf{T}}_1$:

Remember your cross product rules: $\hat{\mathbf{i}}\times \hat{\mathbf{i}}=0$ and $\hat{\mathbf{j}}\times \hat{\mathbf{i}}=-\hat{\mathbf{k}}$. Applying these, we get:

The problem explicitly states that this torque is equal to $\tau \hat{\mathbf{k}}$. By equating the components, we extract a crucial relationship:

Electric Field Along Y-axis

Now, the environment changes. The dipole is subjected to a new electric field, ${\mathbf{E}}_2$, which points along the Y-axis and has a magnitude of $\sqrt{3}E$. So, ${\mathbf{E}}_2=\sqrt{3}E\hat{\mathbf{j}}$. Let's find the new torque, ${\mathbf{T}}_2$:

Again, applying the cross product rules ($\hat{\mathbf{i}}\times \hat{\mathbf{j}}=\hat{\mathbf{k}}$ and $\hat{\mathbf{j}}\times \hat{\mathbf{j}}=0$), we find:

Solving for the Angle

We are given a fascinating constraint: the second torque is the exact negative of the first torque. Mathematically, ${\mathbf{T}}_2=-{\mathbf{T}}_1$. Since ${\mathbf{T}}_1=\tau \hat{\mathbf{k}}$, this means ${\mathbf{T}}_2=-\tau \hat{\mathbf{k}}$.

Let's equate the magnitude of our calculated ${\mathbf{T}}_2$ to $-\tau$:

Now, we substitute the expression for $\tau$ that we derived from the first encounter ($\tau =-pE\sin \theta$):

Notice how beautifully the physics constants $p$ and $E$ cancel out from both sides, leaving us with a pure trigonometric equation:

Dividing both sides by $\cos \theta$, we get:

For an acute angle, the only solution to this equation is:

This problem is a masterclass in why vector notation is superior to scalar formulas. If you had only used the magnitude formula $\tau =pE\sin \theta$, you would have had to manually determine the direction of the torque for both cases using the right-hand rule, which is a common source of sign errors under exam pressure. By trusting the cross product algebra, the negative signs naturally took care of themselves, leading us straight to the correct answer.