Question

An electric dipole is placed at an angle of 30° with an electric field of intensity 2x10⁵ NC-1. It experiences a torque equal to 4 Nm. Calculate the magnitude of charge on the dipole, if the dipole length is 2cm.

• A. 2mC
• B. 8mC
• C. 6mC
• D. 4mC

Answer 1

The Correct Option is A

To solve the given problem, we need to calculate the magnitude of the charge on the dipole using the given information. Let's go through the solution step-by-step:

Identify the given parameters:

• Angle between the dipole and the electric field, \(\theta = 30^\circ\)
• Electric field intensity, \(E = 2 \times 10^5 \, \text{N/C}\)
• Torque experienced by the dipole, \(\tau = 4 \, \text{Nm}\)
• Length of the dipole, \(d = 2 \, \text{cm} = 0.02 \, \text{m}\)

Use the formula for torque on an electric dipole:

The torque \(\tau\) experienced by an electric dipole in a uniform electric field is given by:

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

where \(p\) is the dipole moment, and it is defined as \(p = q \cdot d\), where \(q\) is the charge on the dipole.

Substitute the known values and solve for the charge \(q\):

\(4 = q \cdot 0.02 \cdot (2 \times 10^5) \sin 30^\circ\)

Calculate the sine of the angle:

\(\sin 30^\circ = \frac{1}{2}\)

Simplify and solve for \(q\):

\(4 = q \cdot 0.02 \cdot (2 \times 10^5) \cdot \frac{1}{2}\)

\(4 = q \cdot 0.02 \cdot 10^5\)

\(4 = q \cdot 2000\)

\(q = \frac{4}{2000} = 0.002 \, \text{C}\)

\(0.002 \, \text{C} = 2 \, \text{mC}\)

The magnitude of the charge on the dipole is 2 mC.

Correct Answer: 2 mC