Question

A dipole with two electric charges of $ 2 \, \mu C $ magnitude each, with separation distance $ 0.5 \, \mu m $, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of $ 5 \, V $ is applied. Separation between the plates is $ 0.5 \, mm $. If the dipole is rotated by $ 30^\circ $ from the axis, it tends to realign in the direction due to a torque. The value of torque is :

• A. \( 5 \times 10^{-9} \, Nm \)
• B. \( 5 \times 10^{-3} \, Nm \)
• C. \( 2.5 \times 10^{-12} \, Nm \)
• D. \( 2.5 \times 10^{-9} \, Nm \)

Hint:

Remember the formula for the electric field between parallel plates of a capacitor (\(E = V/D\)) and the torque on an electric dipole in a uniform electric field (\(\tau = p E \sin \theta\)). Ensure consistent SI units throughout the calculation.

Answer 1

The Correct Option is A

To find the torque experienced by the electric dipole when it is rotated by \( 30^\circ \), we first need to understand the relationship between torque (\( \tau \)), electric field (\( E \)), dipole moment (\( p \)), and the angle (\( \theta \)) between the dipole and the electric field.

The torque (\( \tau \)) experienced by a dipole of moment \( \mathbf{p} \) in a uniform electric field \( \mathbf{E} \) is given by the equation:

\(\tau = pE \sin \theta\)

Here, the dipole moment \( p \) is the product of the charge (\( q \)) and the separation distance (\( d \)):

\(p = q \times d\)

Given:

• Charge, \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \)
• Separation distance, \( d = 0.5 \, \mu m = 0.5 \times 10^{-6} \, m \)
• Voltage, \( V = 5 \, V \)
• Separation between capacitor plates, \( s = 0.5 \, mm = 0.5 \times 10^{-3} \, m \)
• Angle, \( \theta = 30^\circ \)

 

First, calculate the electric field (\( E \)) using the voltage and separation:

\(E = \frac{V}{s} = \frac{5}{0.5 \times 10^{-3}} = 10,000 \, V/m\)

Next, calculate the dipole moment:

\(p = q \times d = (2 \times 10^{-6}) \times (0.5 \times 10^{-6}) = 1 \times 10^{-12} \, Cm\)

Now, calculate the torque:

\(\tau = pE \sin \theta = (1 \times 10^{-12}) \times (10,000) \times \sin 30^\circ\)

Since \( \sin 30^\circ = 0.5 \):

\(\tau = (1 \times 10^{-12} \, Cm) \times (10,000 \, V/m) \times 0.5 = 5 \times 10^{-9} \, Nm\)

Therefore, the torque is \(5 \times 10^{-9} \, Nm\), which matches the correct answer option.

Answer 2

Step 1: Identify the given parameters and convert them to SI units.
Magnitude of each charge, \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \) 
Separation distance between the charges (dipole length), \( d = 0.5 \, \mu m = 0.5 \times 10^{-6} \, m \) 
Potential difference across the capacitor plates, \( V = 5 \, V \) 
Separation between the capacitor plates, \( D = 0.5 \, mm = 0.5 \times 10^{-3} \, m \) 
Angle by which the dipole is rotated from the electric field direction, \( \theta = 30^\circ \) 
Step 2: Calculate the electric field between the capacitor plates.
The electric field \( E \) between the plates of a parallel plate capacitor is given by: \[ E = \frac{V}{D} \] 
Substituting the given values: \[ E = \frac{5 \, V}{0.5 \times 10^{-3} \, m} = 10^4 \, V/m \] 
Step 3: Calculate the dipole moment \( p \).
The dipole moment \( p \) is given by the product of the magnitude of one of the charges and the separation distance between the charges: \[ p = q \times d \] Substituting the given values: \[ p = (2 \times 10^{-6} \, C) \times (0.5 \times 10^{-6} \, m) = 1 \times 10^{-12} \, C \cdot m \] 
Step 4: Calculate the torque \( \tau \) on the dipole.
The torque \( \tau \) on an electric dipole placed in a uniform electric field \( E \) at an angle \( \theta \) with the field is given by: \[ \tau = p E \sin \theta \] Substituting the calculated values: \[ \tau = (1 \times 10^{-12} \, C \cdot m) \times (10^4 \, V/m) \times \sin(30^\circ) \] We know that \( \sin(30^\circ) = \frac{1}{2} \). \[ \tau = 1 \times 10^{-8} \times \frac{1}{2} \, N \cdot m = 0.5 \times 10^{-8} \, N \cdot m = 5 \times 10^{-9} \, N \cdot m \] The value of the torque is \( 5 \times 10^{-9} \, Nm \), which corresponds to option (1).