Question

An electric dipole is placed at an angle of $30^\circ$ with an electric field intensity $3 \times 10^5 \text{ NC}^{-1}$ experiences a torque $3 \text{ Nm}$. If the length of dipole is $2 \text{ cm}$, the charge on the dipole is (in milli coloumb)}

• A. 2
• B. 3
• C. 1.5
• D. 1
• E. 4

Hint:

Be careful with the length of the dipole. In the formula $p = q(2l)$, the variable '2l' represents the entire separation between the charges, which is given as 2 cm here.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
An electric dipole placed in a uniform electric field experiences a turning force (torque) that attempts to align it with the field. The torque depends on the field strength, the dipole moment, and the angle.
Step 2: Key Formula or Approach:
Torque on a dipole: \(\tau = pE \sin\theta\).
Dipole moment: \(p = q \times 2a\) (where \(2a\) is the full length of the dipole).
Substitute to form: \(\tau = (q \times 2a) E \sin\theta\).
Solve for \(q\).
Step 3: Detailed Explanation:
Given values:
Angle \(\theta = 30^\circ\)
Electric field \(E = 3 \times 10^5 \text{ N/C}\)
Torque \(\tau = 3 \text{ Nm}\)
Length of dipole \(2a = 2 \text{ cm} = 0.02 \text{ m}\)
Substitute the values into the torque formula:
\[ 3 = q \times (0.02) \times (3 \times 10^5) \times \sin(30^\circ) \] We know \(\sin(30^\circ) = 0.5\).
\[ 3 = q \times 0.02 \times 3 \times 10^5 \times 0.5 \] Cancel the 3 from both sides:
\[ 1 = q \times 0.02 \times 10^5 \times 0.5 \] \[ 1 = q \times 0.01 \times 10^5 \] \[ 1 = q \times 10^3 \] Solve for \(q\):
\[ q = 10^{-3} \text{ C} \] Convert Coulombs to milli-Coulombs (mC):
\[ q = 1 \text{ mC} \] Step 4: Final Answer:
The charge on the dipole is 1.