Question

An electric dipole of length \(0.5 μ\text{m}\) is placed with its axis making an angle of \(30^{\circ}\) with uniform electric field \(10^4 \text{V/m}\). If it experiences a torque of \(5\times 10^{-9} \text{Nm}\), the magnitude of the charge on the dipole is (\(sin30^{\circ} = 0.5\))

• A. \(1 μ\text{C}\)
• B. \(1.5 μ\text{C}\)
• C. \(2 μ\text{C}\)
• D. \(2.5 μ\text{C}\)

Hint:

Torque is p E sin theta, with dipole moment p equal to q times length.

Answer 1

The Correct Option is C

Step 1: Formula
\(\tau = pE\sin\theta = qLE\sin\theta\).

Step 2: Solve for q
\[ q = \frac{\tau}{LE\sin\theta} = \frac{5\times10^{-9}}{(0.5\times10^{-6})(10^4)(0.5)} = \frac{5\times10^{-9}}{2.5\times10^{-3}} \]

Step 3: Result
\(q = 2\times10^{-6}\ \text{C} = 2\ \mu\text{C}\). Option (C).

Final Answer:
The charge is 2 microcoulomb. \[ \boxed{\text{(C)}\ 2\ \mu\text{C}} \]