Question

An electric dipole of dipole moment is 6.0 x 10^-6 Cm placed in a uniform electric field of 1.5 x 10³ NC^-1 in such a way that dipole moment is along electric field. The work done in rotating dipole by 180º in this field will be _____mJ.

Hint:

Work done depends on the angle of rotation and the strength of the electric field.

Answer 1

Correct Answer: 18

An electric dipole placed in a uniform electric field will experience torque, and work is done on it when it is rotated. The work done \( W \) in rotating a dipole by an angle \( \theta \) in a uniform electric field can be calculated using the formula:

\[ W = pE(\cos \theta_1 - \cos \theta_2) \]

Where:

• p is the dipole moment (6.0 × 10^-6 Cm)
• E is the electric field strength (1.5 × 10³ N/C)
• \(\theta_1\) and \(\theta_2\) are the initial and final angles, respectively.

Initially, the dipole moment is aligned with the field, so \(\theta_1 = 0^\circ\), thus \(\cos \theta_1 = \cos 0^\circ = 1\). After rotation by 180°, \(\theta_2 = 180^\circ\) and \(\cos \theta_2 = \cos 180^\circ = -1\). Substitute these values into the equation:

\[ W = (6.0 \times 10^{-6} \text{ Cm})(1.5 \times 10^3 \text{ N/C})(1 - (-1)) \]

Calculate the expression inside the parenthesis:

\[ 1 - (-1) = 2 \]

Now calculate the work done:

\[ W = (6.0 \times 10^{-6} \text{ Cm})(1.5 \times 10^3 \text{ N/C})(2) = 18 \times 10^{-3} \text{ J} = 18 \text{ mJ} \]

The calculated work done is 18 mJ, which falls within the expected range of 18,18 mJ. Thus, the solution is confirmed to be correct.