Question 286: Electrostatics & Capacitance

Question

Two charges each of magnitude $$0.01 ~\mathrm{C}$$ and separated by a distance of $$0.4 \mathrm{~mm}$$ constitute an electric dipole. If the dipole is placed in an uniform electric field '$$\vec{E}$$' of 10 dyne/C making $$30^{\circ}$$ angle with $$\vec{E}$$, the magnitude of torque acting on dipole is:

Accepted Answer

Given two charges each of magnitude $$0.01 \,	ext{C}$$ and separated by a distance of $$0.4 \,	ext{mm} = 0.4 	imes 10^{-3} \,	ext{m}$$, we can calculate the dipole moment $$\vec{p}$$:
$$\vec{p} = q \cdot \vec{d} = (0.01 \,	ext{C})(0.4 	imes 10^{-3} \,	ext{m}) = 4 	imes 10^{-6} \,	ext{Cm}$$
The dipole is placed in a uniform electric field $$\vec{E}$$ of magnitude $$10 \,	ext{dyne/C}$$, which is equivalent to $$10 	imes 10^{-5} \,	ext{N/C}$$.
The angle between the dipole moment and the electric field is $$30^{\circ}$$.
The torque acting on the dipole can be calculated using the formula:
$$	au = pE \sin{	heta}$$
Substituting the given values:
$$	au = (4 	imes 10^{-6} \,	ext{Cm})(10 	imes 10^{-5} \,	ext{N/C})\sin{30^{\circ}}$$
$$	au = (4 	imes 10^{-6} \,	ext{Cm})(10^{-4} \,	ext{N/C})(\frac{1}{2})$$
$$	au = 2 	imes 10^{-10} \,	ext{Nm}$$
The magnitude of the torque acting on the dipole is $$2.0 	imes 10^{-10} \,	ext{Nm}$$.

Answer

$$4 \cdot 0 	imes 10^{-10} ~\mathrm{Nm}$$

Answer

$$1.5 	imes 10^{-9} ~\mathrm{Nm}$$

Answer

$$1.0 	imes 10^{-8} ~\mathrm{Nm}$$

Answer

$$2.0 	imes 10^{-10} ~\mathrm{Nm}$$

Electrostatics & Capacitance

Two charges each of magnitude $0.01 ~\mathrm{C} and separated by a distance of 0.4 \mathrm{~mm} constitute an electric dipole. If the dipole is placed in an uniform electric field '\vec{E}' of 10 dyne/C making 30^{\circ} angle with \vec{E}$, the magnitude of torque acting on dipole is:

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