Question

An electric dipole is placed at an angle of $30^{\circ}$ with an electric field intensity $2 \times 10^{5} N / C .$ It experiences a torque equal to $4\, Nm$. The charge on the dipole, if the dipole length is $2\, cm$, is

• A. $8\,mC$
• B. $4\,mC$
• C. $6\,mC$
• D. $2 \,mC$

Answer 1

The Correct Option is D

To find the charge on the dipole, we need to use the relationship between torque, electric field, dipole moment, and the angle at which the dipole is placed in the electric field. The formula for torque (\tau) experienced by an electric dipole in a uniform electric field is given by:

\tau = pE\sin\theta

where:

• \tau = 4 \, \text{Nm} is the torque experienced by the dipole.
• E = 2 \times 10^{5} \, \text{N/C} is the electric field intensity.
• \theta = 30^{\circ} is the angle between the dipole and the electric field.
• p is the dipole moment, which is the product of the charge q on the dipole and the dipole length d.

From the formula, we can express the dipole moment (p) as:

p = q \times d

Using the torque expression:

\tau = q \times d \times E \times \sin\theta

Rearranging for q, we get:

q = \frac{\tau}{d \times E \times \sin\theta}

For this problem:

• The dipole length d = 2\, \text{cm} = 0.02\, \text{m}.
• \sin 30^{\circ} = 0.5.

Substituting these values into the equation for q:

q = \frac{4}{0.02 \times 2 \times 10^{5} \times 0.5} = \frac{4}{2000} = 0.002 \, \text{C}

Converting to milli-Coulombs (mC):

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

Therefore, the charge on the dipole is 2 \, \text{mC}, which matches the correct option.