Question 713: Electrostatics & Capacitance

Question

An electric dipole is placed at an angle of $$30^{\circ}$$ with an electric field of intensity $$2 	imes 10^{5} \mathrm{NC}^{-1}$$. It experiences a torque equal to $$4 ~\mathrm{N~m}$$. Calculate the magnitude of charge on the dipole, if the dipole length is $$2 \mathrm{~cm}$$.

Accepted Answer

The torque ÃÂ experienced by an electric dipole in an electric field is given by the formula:
$$	au = pE\sin{	heta}$$
where p is the electric dipole moment, E is the electric field intensity, and ÃÂ¸ is the angle between the dipole and the electric field. The electric dipole moment p can be expressed as:
$$p = qd$$
where q is the charge on the dipole, and d is the dipole length.
We are given the following values:
Torque ÃÂ = 4 NÃÂ·m
Electric field intensity E = $$2 	imes 10^5 ~\mathrm{NC}^{-1}$$
Angle ÃÂ¸ = $$30^{\circ}$$
Dipole length d = 2 cm = 0.02 m
We need to find the charge q on the dipole. Let's first solve for the electric dipole moment p:
$$	au = pE\sin{	heta}$$
$$ \Rightarrow $$ $$p = \frac{	au}{E\sin{	heta}}$$
Substituting the given values:
$$p = \frac{4}{(2 	imes 10^5) \sin{30^{\circ}}} = \frac{4}{(2 	imes 10^5)(0.5)} = \frac{4}{10^5} = 4 	imes 10^{-5} ~\mathrm{C~m}$$
Now, let's solve for the charge q using the formula:
$$ \Rightarrow $$ $$p = qd$$
$$q = \frac{p}{d}$$
Substituting the values for p and d:
$$q = \frac{4 	imes 10^{-5}}{0.02} = 2 	imes 10^{-3} ~\mathrm{C} = 2 ~\mathrm{mC}$$
So, the magnitude of the charge on the dipole is 2 mC.

Answer

6 mC

Answer

4 mC

Answer

2 mC

Answer

8 mC

Electrostatics & Capacitance

An electric dipole is placed at an angle of $30^{\circ} with an electric field of intensity 2 \times 10^{5} \mathrm{NC}^{-1}. It experiences a torque equal to 4 ~\mathrm{N~m}. Calculate the magnitude of charge on the dipole, if the dipole length is 2 \mathrm{~cm}$.

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