Question 759: Electrostats & Capacitance

Question

An electric dipole of length 2 cm is placed with its axis making an angle of $60^{\circ}$ to a uniform electric field of $10^{+5} \mathrm{~N} / \mathrm{C}$. If it experiences a torque of $9 \sqrt{3} \mathrm{Nm}$, the magnitude of the charge on the dipole is $\left(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$

Accepted Answer

Step 1: Write the formula for torque on an electric dipole:
$$ 	ext{Torque} = p E \sin	heta $$
Here, $p$ is the dipole moment, $E$ is the electric field, and $	heta$ is the angle between the dipole and field.
Step 2: Replace $p$ with $ql$:
$p = q \cdot l$, where $q$ is the charge and $l$ is the length of the dipole.
So,
$$ 	ext{Torque} = q \cdot l \cdot E \cdot \sin 60^\circ $$
Step 3: Plug in the values given in the problem:
Torque $= 9\sqrt{3}$ Nm, $l = 2$ cm $= 2 	imes 10^{-2}$ m, $E = 10^5$ N/C, $\sin 60^\circ = \frac{\sqrt{3}}{2}$
$$ 9\sqrt{3} = q 	imes 2 	imes 10^{-2} 	imes 10^{5} 	imes \frac{\sqrt{3}}{2} $$
Step 4: Simplify and solve for $q$:
The $\sqrt{3}$ on both sides cancels out.
$$ 9 = q 	imes 2 	imes 10^{-2} 	imes 10^{5} 	imes \frac{1}{2} $$
Calculate $2 	imes 10^{-2} 	imes 10^{5} = 2 	imes 10^{3}$.
So,
$$ 9 = q 	imes 2 	imes 10^{3} 	imes \frac{1}{2} $$
The $2$ and $\frac{1}{2}$ also cancel, so:
$$ 9 = q 	imes 10^{3} $$
Step 5: Calculate $q$:
$$ q = \frac{9}{10^{3}} = 9 	imes 10^{-3} \mathrm{C} $$
Final Answer:
The magnitude of the charge on the dipole is $9 	imes 10^{-3}$ C.

Answer

$7 	imes 10^{-3} \mathrm{C}$

Answer

$8 	imes 10^{-3} \mathrm{C}$

Answer

$9 	imes 10^{-3} \mathrm{C}$

Answer

$\frac{9}{2} 	imes 10^{-3} \mathrm{C}$

Electrostats & Capacitance

An electric dipole of length 2 cm is placed with its axis making an angle of 60^{\circ} to a uniform electric field of 10^{+5} \mathrm{~N} / \mathrm{C}. If it experiences a torque of 9 \sqrt{3} \mathrm{Nm}, the magnitude of the charge on the dipole is \left(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)

Choose the correct answer: