Question

\( 1\,\mu\text{C} \), \( -1\,\mu\text{C} \) charges are placed at a distance of 5 cm in forming a dipole. The amount of torque required to place this dipole perpendicular to an electric field of \( 3\times10^{5}~\text{NC}^{-1} \) is given by:

• A. \( 5\times10^{-3}~\text{N}\cdot\text{m} \)
• B. \( 15\times10^{-3}~\text{N}\cdot\text{m} \)
• C. \( 1\times10^{-3}~\text{N}\cdot\text{m} \)
• D. \( 10\times10^{-3}~\text{N}\cdot\text{m} \)

Hint:

The torque acting on an electric dipole reaches its maximum possible value when it is oriented perpendicular (\( 90^\circ \)) to the electric field lines, and drops to zero when it aligns parallel (\( 0^\circ \)) or antiparallel (\( 180^\circ \)) to the field.

Answer 1

The Correct Option is B

Step 1: Recall the dipole moment.
An electric dipole is two equal and opposite charges a small distance apart. Its dipole moment is one charge times the separation: \[ p = q\,d \]
Step 2: Recall the torque formula.
In a uniform field $E$, the torque on a dipole tilted by angle $\theta$ to the field is: \[ \tau = pE\sin\theta \]
Step 3: Find the dipole moment.
With $q = 10^{-6}$ C and $d = 0.05$ m: \[ p = (10^{-6})(5\times10^{-2}) = 5\times10^{-8}\ \text{C}\,\text{m} \]
Step 4: Set the angle.
To make the dipole perpendicular to the field, $\theta = 90^{\circ}$, so $\sin 90^{\circ} = 1$. This is also the maximum torque case.
Step 5: Put in the field.
With $E = 3\times10^{5}\ \text{NC}^{-1}$: \[ \tau = (5\times10^{-8})(3\times10^{5})(1) \]
Step 6: Compute.
\[ \tau = 15\times10^{-3}\ \text{N}\,\text{m} \] \[ \boxed{\tau = 15\times10^{-3}\ \text{N}\,\text{m}} \]