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Step-by-Step Solution
Step 1: Write the expression for torque on an electric dipole
An electric dipole with dipole moment vector $\overrightarrow{p}$ placed in an electric field $\overrightarrow{E}$ experiences a torque given by:
$$
\overrightarrow{\tau} = \overrightarrow{p} \times \overrightarrow{E}.
$$
In magnitude form, this can also be written as:
$$
\tau = pE \sin \theta,
$$
where $\theta$ is the angle between $\overrightarrow{p}$ and $\overrightarrow{E}$.
Step 2: Represent the dipole moment vector in component form
Given that the dipole moment $\overrightarrow{p}$ makes an angle $\theta$ with the x-axis, we can write:
$$
\overrightarrow{p} = p \cos \theta \,\widehat{i} + p \sin \theta \,\widehat{j}.
$$
Step 3: Compute the torque in the first electric field $\overrightarrow{E}_1$
The first electric field is:
$$
\overrightarrow{E}_1 = E\,\widehat{i}.
$$
Using $\overrightarrow{\tau}_1 = \overrightarrow{p} \times \overrightarrow{E}_1$, we calculate:
$$
\overrightarrow{\tau}_1
= \bigl(p \cos \theta \,\widehat{i} + p \sin \theta \,\widehat{j}\bigr)
\times \bigl(E\,\widehat{i}\bigr).
$$
The cross product of $\widehat{i}$ with itself is zero, and the cross product of $\widehat{j}$ with $\widehat{i}$ gives $-\widehat{k}$. Hence,
$$
\overrightarrow{\tau}_1
= pE \sin \theta \,\bigl(-\widehat{k}\bigr).
$$
So in vector form:
$$
\overrightarrow{\tau}_1 = -\,pE \sin\theta \,\widehat{k}.
$$
Its magnitude is $\tau_1 = pE \sin \theta$ along the negative $z$-direction.
Step 4: Compute the torque in the second electric field $\overrightarrow{E}_2$
The second electric field is:
$$
\overrightarrow{E}_2 = \sqrt{3}\,E\,\widehat{j}.
$$
Using $\overrightarrow{\tau}_2 = \overrightarrow{p} \times \overrightarrow{E}_2$, we get:
$$
\overrightarrow{\tau}_2
= \bigl(p \cos \theta \,\widehat{i} + p \sin \theta \,\widehat{j}\bigr)
\times \bigl(\sqrt{3}E\,\widehat{j}\bigr).
$$
Here, $\widehat{j} \times \widehat{j}$ is zero, and $\widehat{i} \times \widehat{j}$ gives $+\widehat{k}$. Hence,
$$
\overrightarrow{\tau}_2
= \sqrt{3}\,pE \cos \theta \,\widehat{k}.
$$
Its magnitude is $\tau_2 = \sqrt{3} \, pE \cos \theta$ along the positive $z$-direction.
Step 5: Apply the given condition $\overrightarrow{\tau}_2 = -\,\overrightarrow{\tau}_1$
The problem states that $\overrightarrow{\tau}_2$ is equal to $-\,\overrightarrow{\tau}_1$. In vector form, this means:
$$
\sqrt{3}\,pE \cos \theta \,\widehat{k}
= -\bigl(-\,pE \sin \theta \,\widehat{k}\bigr).
$$
That simplifies to:
$$
\sqrt{3}\,pE \cos \theta \,\widehat{k}
= pE \sin \theta \,\widehat{k}.
$$
Step 6: Solve for $\theta$
Since the $\widehat{k}$ directions match, we can compare coefficients:
$$
\sqrt{3}\,\cos \theta = \sin \theta.
$$
Dividing both sides by $\cos\theta$ (assuming $\cos\theta \neq 0$), we get:
$$
\sqrt{3} = \tan \theta.
$$
Hence,
$$
\tan \theta = \sqrt{3}
\quad\Longrightarrow\quad
\theta = 60^\circ.
$$
Answer: 60°
