Question 353: Electrostats & Capacitance

Question

Two point charges $-4 \mu \mathrm{c}$ and $4 \mu \mathrm{c}$, constituting an electric dipole, are placed at $(-9,0,0) \mathrm{cm}$ and $(9,0,0) \mathrm{cm}$ in a uniform electric field of strength $10^4 \mathrm{NC}^{-1}$. The work done on the dipole in rotating it from the equilibrium through $180^{\circ}$ is :

Accepted Answer

The work done on an electric dipole in a uniform electric field when it is rotated from an angle $	heta_1$ to an angle $	heta_2$ is given by:
$$ W = -pE(\cos 	heta_2 - \cos 	heta_1) $$
where:
$ p = q \cdot d $ is the dipole moment,
$ E $ is the electric field strength,
$ 	heta_1 $ and $ 	heta_2 $ are the initial and final angles between the dipole moment and the electric field.
Given:
Charge $ q = 4 \, \mu \mathrm{C} = 4 	imes 10^{-6} \, \mathrm{C} $
Distance $ d = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m} $
Electric field $ E = 10^4 \, \mathrm{N/C} $
Initial angle $	heta_1 = 0^\circ$ (aligned with the field, equilibrium position)
Final angle $	heta_2 = 180^\circ$
First, calculate the dipole moment $ p $:
$$ p = q \cdot d = 4 	imes 10^{-6} \, \mathrm{C} \cdot 0.18 \, \mathrm{m} = 7.2 	imes 10^{-7} \, \mathrm{Cm} $$
Calculate the work done $ W $:
$$ W = -pE(\cos 180^\circ - \cos 0^\circ) $$
$$ W = -7.2 	imes 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot (-1 - 1) $$
$$ W = 7.2 	imes 10^{-7} \, \mathrm{Cm} \cdot 10^4 \, \mathrm{N/C} \cdot 2 $$
$$ W = 1.44 	imes 10^{-2} \, \mathrm{J} = 14.4 \, \mathrm{mJ} $$
Therefore, the work done on the dipole in rotating it from the equilibrium position through $180^\circ$ is
14.4 mJ
(Option D).

Answer

18.4 mJ

Answer

12.4 mJ

Answer

16.4 mJ

Answer

14.4 mJ

Electrostats & Capacitance

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