Two point charges – 4 $\mathrm{\mu }$C and 4 $\mathrm{\mu }$C, constituting an electric dipole, are placed at (– 9, 0, 0) cm and (9, 0, 0) cm in a uniform electric field of strength 10⁴ NC^–1. The work done on the dipole in rotating it from the equilibrium through 180° is :

To find the work done in rotating the electric dipole from equilibrium (0°) to 180°, we use the potential energy formula of a dipole in a uniform electric field:

$$U = - pE \cos \theta$$

where:

$p = q \cdot d$ is the dipole moment,

$E$ is the electric field strength,

$\theta$ is the angle between $p$ and $E$.

Step 1: Given Data

Charge: $q = 4 \ \mu C = 4 \times 10^{-6} C$

Distance between charges: $d = 9 - (-9) = 18$ cm = 0.18 m

Electric field: $E = 10^4$ N/C

Initial angle: $\theta_1 = 0^\circ$ (equilibrium position)

Final angle: $\theta_2 = 180^\circ$

Step 2: Dipole Moment Calculation

$$p = q \cdot d = (4 \times 10^{-6}) \times (0.18)$$

 $p=7.2\times {10}^{-7}\text{C·m}$

Step 3: Work Done Calculation

Work done in rotating the dipole from $\theta_1$ to $\theta_2$ is given by:

$$W = U_2 - U_1$$

 $$W = (- pE \cos \theta_2) - (- pE \cos \theta_1)$$ $$W = pE (\cos 0^\circ - \cos 180^\circ)$$

Since $\cos 0^\circ = 1$ and $\cos 180^\circ = -1$,

$$W = pE (1 - (-1))$$

 $$W = 2 pE$$

Substituting values:

$$W = 2 \times (7.2 \times 10^{-7}) \times (10^4)$$

 $$W = 2 \times 7.2 \times 10^{-3}$$ $$W = 0.0144 \ \text{J} = 14.4 \ \text{mJ}$$

Final Answer:

The work done is 14.4 mJ.