Question

An electric dipole is placed parallel to the electric field . Let 'W' be the work done in rotating the dipole by \(60^{\circ }\) . Now, it is further rotated by \(120^{\circ }\), the total work done is
\([\sin 30^{\circ }=\cos 60^{\circ }=0.5,\ \cos 0^{\circ }=\sin 90^{\circ }=1,\ \cos \pi ^{c}=-1]\)

• A. \(\text{W}/2\)
• B. \(2\text{W}\)
• C. \(3\text{W}\)
• D. \(4\text{W}\)

Hint:

Use work equals pE times the change in cosine of the angle from the starting orientation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Work done in turning a dipole from angle \(\theta_1\) to \(\theta_2\) is \(W=pE(\cos\theta_1-\cos\theta_2)\).

Step 2: First rotation
From \(0^{\circ}\) to \(60^{\circ}\): \(W=pE(1-0.5)=\dfrac{pE}{2}\), so \(pE=2W\).

Step 3: Total rotation
A further \(120^{\circ}\) takes it to \(180^{\circ}\). Total work from the start \(=pE(\cos0^{\circ}-\cos180^{\circ})=2pE\).

Step 4: Result
Total work \(=2\times2W=4W\)

Final Answer:
The total work done is \(4\text{W}\). \[ \boxed{\text{(D) }4\text{W}} \]