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 find the magnitude of the charge on the dipole, we can use the formula for the torque experienced by a dipole in an electric field:

τ = pEsinθ

where:

• τ is the torque (4 Nm),
• p is the dipole moment,
• E is the electric field intensity (2×10⁵ N/C),
• θ is the angle between the dipole and the electric field (30°).

The dipole moment p is given by:

p = qd

where:

• q is the charge,
• d is the dipole length (2 cm = 0.02 m).

First, express the sine of the angle:

sin30° = 0.5

Substitute the known values into the torque formula:

4 = q × 0.02 × 2×10⁵ × 0.5

Simplify the equation:

4 = q × 0.01 × 2×10⁵

4 = q × 2000

Solve for q:

q = 4 / 2000 = 0.002 C = 2 mC

Therefore, the magnitude of the charge on the dipole is 2 mC.

Answer 2

The correct option is (C): 2mC

Here \(\theta =30^{\circ}\), \(E=2\times10^{5}NC^{-1}\)

\(\tau = 4\,Nm\),\(l=2cm=0.02m.\)

\(\tau=pEsin\theta=(ql)E\,sin\theta\)

\(\therefore \,\,q=\frac{\tau}{El\,sin\theta}=\frac{4}{2\times 10^5\times 0.02\times\frac{1}{2}}\)

\(\therefore \frac{4}{2\times10^3}=2\times10^{-3}C=2mC\)