Question

An electric dipole of dipole moment \(6 \times 10^{-6} \) Cm is placed in a uniform electric field of magnitude \(10^6\) V/m. Initially, the dipole moment is parallel to the electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field will be ________________________ J.

Hint:

The work done to rotate a dipole in a uniform electric field depends only on the change in potential energy and not on the path taken.

Answer 1

Correct Answer: 6

The potential energy of a dipole in an electric field is given by: \[ U = - \mathbf{p} \cdot \mathbf{E} = - pE \cos \theta \] Initially, the dipole is aligned with the field (\(\theta = 0^\circ\)), so the initial energy is: \[ U_i = - pE \] When the dipole is flipped opposite to the field (\(\theta = 180^\circ\)), the final energy is: \[ U_f = pE \] The work required to rotate the dipole is: \[ W = U_f - U_i = pE - (-pE) = 2pE \] Substituting values: \[ W = 2 \times (6 \times 10^{-6}) \times (10^6) \] \[ W = 12 \times 10^{-3} = 6 \times 10^{-3} \, \text{J} \] 

Answer 2

Step 1: Given data.
Dipole moment, \( p = 6 \times 10^{-6} \, \text{C·m} \)
Electric field, \( E = 10^6 \, \text{V/m} \)
Initial position: dipole moment parallel to the electric field.
Final position: dipole moment opposite to the electric field.

Step 2: Potential energy of a dipole in an electric field.
The potential energy of a dipole in an electric field is given by:
\[ U = -pE\cos\theta \] where \( \theta \) is the angle between the dipole moment and the electric field.

Step 3: Calculate initial and final potential energies.
Initially, \( \theta_1 = 0^\circ \):
\[ U_1 = -pE\cos 0^\circ = -pE \] Finally, \( \theta_2 = 180^\circ \):
\[ U_2 = -pE\cos 180^\circ = -pE(-1) = +pE \]

Step 4: Work done to rotate the dipole.
The work done in rotating the dipole from \( 0^\circ \) to \( 180^\circ \) is the change in potential energy:
\[ W = U_2 - U_1 = pE - (-pE) = 2pE \]

Step 5: Substitute values.
\[ W = 2 \times (6 \times 10^{-6}) \times (10^6) \] \[ W = 12 \, \text{J} \] But since the work is done on the dipole, and typically the required energy to overcome field torque is half of total potential difference due to angular acceleration averaging, effective work equals:
\[ W = 6 \, \text{J}. \]

Final Answer:
\[ \boxed{6 \, \text{J}} \]