Two point charges $+2\ \mu ext{C}$ and $-2\ \mu ext{C}$ are placed $3 ext{ cm}$ apart, forming an electric dipole. This dipole is placed in a uniform electric field of magnitude $2 imes 10^5 ext{ N/C}$ making an angle of $30^\circ$ with the direction of the field. Calculate: The magnitude of the electric dipole moment. The torque acting on the electric dipole. The work done in rotating the dipole from stable equilibrium ($ heta = 0^\circ$) to unstable equilibrium ($ heta = 180^\circ$). The potential energy of the dipole at its present position ($ heta = 30^\circ$).

Q: Two point charges $+2\ \mu\text{C}$ and $-2\ \mu\text{C}$ are placed $3\text{ cm}$ apart, forming an electric dipole. This dipole is placed in a uniform electric field of magnitude $2 \times 10^5\text{ N/C}$ making an angle of $30^\circ$ with the direction of the field. Calculate: The magnitude of the electric dipole moment. The torque acting on the electric dipole. The work done in rotating the dipole from stable equilibrium ($\theta = 0^\circ$) to unstable equilibrium ($\theta = 180^\circ$). The potential energy of the dipole at its present position ($\theta = 30^\circ$).

Given Data: Magnitude of charge, $q = 2\ \mu\text{C} = 2 \times 10^{-6}\text{ C}$ Distance between charges (length of dipole), $2l = 3\text{ cm} = 3 \times 10^{-2}\text{ m}$ Uniform electric field, $E = 2 \times 10^5\text{ N/C}$ Angle with electric field, $\theta = 30^\circ$ --- Step-by-Step Solutions: Electric Dipole Moment ($p$) Formula: $p = q \times (2l)$ $$p = (2 \times 10^{-6}\text{ C}) \times (3 \times 10^{-2}\text{ m})$$ $$p = 6 \times 10^{-8}\text{ C}\cdot\text{m}$$ Answer (1): $6 \times 10^{-8}\text{ C}\cdot\text{m}$ --- Torque Acting on the Dipole ($\tau$) Formula: $\tau = p E \sin\theta$ $$\tau = (6 \times 10^{-8}\text{ C}\cdot\text{m}) \times (2 \times 10^5\text{ N/C}) \times \sin(30^\circ)$$ $$\text{Since } \sin 30^\circ = 0.5:$$ $$\tau = (12 \times 10^{-3}) \times 0.5$$ $$\tau = 6 \times 10^{-3}\text{ N}\cdot\text{m}$$ Answer (2): $6 \times 10^{-3}\text{ N}\cdot\text{m}$ --- Work Done in Rotating from $\theta1 = 0^\circ$ to $\theta2 = 180^\circ$ ($W$) Formula: $W = p E (\cos\theta1 - \cos\theta2)$ $$W = p E (\cos 0^\circ - \cos 180^\circ)$$ $$\text{Since } \cos 0^\circ = 1 \text{ and } \cos 180^\circ = -1:$$ $$W = p E 1 - (-1) = 2 p E$$ $$W = 2 \times (6 \times 10^{-8}\text{ C}\cdot\text{m}) \times (2 \times 10^5\text{ N/C})$$ $$W = 2 \times (1.2 \times 10^{-2})$$ $$W = 2.4 \times 10^{-2}\text{ J} = 0.024\text{ J}$$ Answer (3): $2.4 \times 10^{-2}\text{ J}$ (or $0.024\text{ J}$) --- Potential Energy of Dipole at $\theta = 30^\circ$ ($U$) Formula: $U = -p E \cos\theta$ $$U = -(6 \times 10^{-8}\text{ C}\cdot\text{m}) \times (2 \times 10^5\text{ N/C}) \times \cos(30^\circ)$$ $$\text{Since } \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866:$$ $$U = -(1.2 \times 10^{-2}) \times 0.866$$ $$U = -1.039 \times 10^{-2}\text{ J}$$ Answer (4): $-1.039 \times 10^{-2}\text{ J}$

Question

Two point charges $+2\ \mu	ext{C}$ and $-2\ \mu	ext{C}$ are placed $3	ext{ cm}$ apart, forming an electric dipole. This dipole is placed in a uniform electric field of magnitude $2 	imes 10^5	ext{ N/C}$ making an angle of $30^\circ$ with the direction of the field. Calculate: The magnitude of the electric dipole moment. The torque acting on the electric dipole. The work done in rotating the dipole from stable equilibrium ($	heta = 0^\circ$) to unstable equilibrium ($	heta = 180^\circ$). The potential energy of the dipole at its present position ($	heta = 30^\circ$).

Accepted Answer

Given Data: Magnitude of charge, $q = 2\ \mu	ext{C} = 2 	imes 10^{-6}	ext{ C}$ Distance between charges (length of dipole), $2l = 3	ext{ cm} = 3 	imes 10^{-2}	ext{ m}$ Uniform electric field, $E = 2 	imes 10^5	ext{ N/C}$ Angle with electric field, $	heta = 30^\circ$ --- Step-by-Step Solutions: Electric Dipole Moment ($p$) Formula: $p = q 	imes (2l)$ $$p = (2 	imes 10^{-6}	ext{ C}) 	imes (3 	imes 10^{-2}	ext{ m})$$ $$p = 6 	imes 10^{-8}	ext{ C}\cdot	ext{m}$$ Answer (1): $6 	imes 10^{-8}	ext{ C}\cdot	ext{m}$ --- Torque Acting on the Dipole ($	au$) Formula: $	au = p E \sin	heta$ $$	au = (6 	imes 10^{-8}	ext{ C}\cdot	ext{m}) 	imes (2 	imes 10^5	ext{ N/C}) 	imes \sin(30^\circ)$$ $$	ext{Since } \sin 30^\circ = 0.5:$$ $$	au = (12 	imes 10^{-3}) 	imes 0.5$$ $$	au = 6 	imes 10^{-3}	ext{ N}\cdot	ext{m}$$ Answer (2): $6 	imes 10^{-3}	ext{ N}\cdot	ext{m}$ --- Work Done in Rotating from $	heta1 = 0^\circ$ to $	heta2 = 180^\circ$ ($W$) Formula: $W = p E (\cos	heta1 - \cos	heta2)$ $$W = p E (\cos 0^\circ - \cos 180^\circ)$$ $$	ext{Since } \cos 0^\circ = 1 	ext{ and } \cos 180^\circ = -1:$$ $$W = p E 1 - (-1) = 2 p E$$ $$W = 2 	imes (6 	imes 10^{-8}	ext{ C}\cdot	ext{m}) 	imes (2 	imes 10^5	ext{ N/C})$$ $$W = 2 	imes (1.2 	imes 10^{-2})$$ $$W = 2.4 	imes 10^{-2}	ext{ J} = 0.024	ext{ J}$$ Answer (3): $2.4 	imes 10^{-2}	ext{ J}$ (or $0.024	ext{ J}$) --- Potential Energy of Dipole at $	heta = 30^\circ$ ($U$) Formula: $U = -p E \cos	heta$ $$U = -(6 	imes 10^{-8}	ext{ C}\cdot	ext{m}) 	imes (2 	imes 10^5	ext{ N/C}) 	imes \cos(30^\circ)$$ $$	ext{Since } \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866:$$ $$U = -(1.2 	imes 10^{-2}) 	imes 0.866$$ $$U = -1.039 	imes 10^{-2}	ext{ J}$$ Answer (4): $-1.039 	imes 10^{-2}	ext{ J}$

Given Data:

Step-by-Step Solutions:

1. Electric Dipole Moment ($p$)

2. Torque Acting on the Dipole ($\tau$)

3. Work Done in Rotating from $\theta_1 = 0^\circ$ to $\theta_2 = 180^\circ$ ($W$)

4. Potential Energy of Dipole at $\theta = 30^\circ$ ($U$)