Question

Two point charges \( +10\,\mu C \) and \( -10\,\mu C \) are situated 2 cm apart. This dipole is placed in a uniform electric field of \( 1\times10^{5} \) volt/metre. Calculate the energy required to rotate the dipole from the position of \( 0^\circ \) to \( 180^\circ \).

Hint:

Find the dipole moment p = q(2a), then use W = pE(cos0 − cos180) = 2pE for a rotation from 0 to 180 degrees.

Answer 1

Step 1: List the data.
Charge \( q = 10\,\mu C = 10\times10^{-6}\,C = 1\times10^{-5}\,C \)
Separation \( 2a = 2\,cm = 0.02\,m \)
Electric field \( E = 1\times10^{5}\,V/m \)
Initial angle \( \theta_1 = 0^\circ \), final angle \( \theta_2 = 180^\circ \)

Step 2: Dipole moment.
\[ p = q \times (2a) = (1\times10^{-5})\times(0.02) = 2\times10^{-7}\ \text{C·m} \]

Step 3: Formula for work to rotate a dipole.
The work done to rotate a dipole from angle \( \theta_1 \) to \( \theta_2 \) in a uniform field is
\[ W = pE\,(\cos\theta_1 - \cos\theta_2) \]

Step 4: Substitute.
\[ W = pE\,(\cos 0^\circ - \cos 180^\circ) = pE\,[\,1 - (-1)\,] = 2pE \]
\[ W = 2 \times (2\times10^{-7}) \times (1\times10^{5}) \]

Step 5: Arithmetic.
\[ W = 2 \times 2\times10^{-2} = 4\times10^{-2}\ \text{J} = 0.04\ \text{J} \]
\[\boxed{W = 4\times10^{-2}\ \text{J} = 0.04\ \text{J}}\]