Question

Two small spherical balls of mass 10 g each with charges $-2 \mu \mathrm{C}$ and $2 \mu \mathrm{C}$, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of $100 \mu \mathrm{C} / \mathrm{m}^{2}$ such that length of rod makes an angle of $30^{\circ}$ with electric field generated by charge sheet. Net torque acting on the rod is:

• A. 112 Nm
• B. 1.12 Nm
• C. 2.24 Nm
• D. 11.2 Nm

Hint:

The torque acting on a dipole in an electric field is given by the product of the dipole moment and the electric field.

Answer 1

The Correct Option is B

1. Electric field due to the charge sheet: \[ E = \frac{\sigma}{2 \epsilon_0} \]
2. Torque acting on the rod: \[ \tau = PE \sin \theta \] \[ \tau = \left[ 2 \times 10^{-6} \times \frac{2}{10} \right] \left[ \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \right] \frac{1}{2} \] \[ \tau = \frac{10}{8.85} = 1.12 \mathrm{~Nm} \] Therefore, the correct answer is (2) 1.12 Nm.

Answer 2

We are given two small spherical balls of equal mass \( m = 10 \, \mathrm{g} \) carrying charges \( q_1 = -2 \, \mu\mathrm{C} \) and \( q_2 = +2 \, \mu\mathrm{C} \). These are attached to the ends of a light rigid rod of length \( L = 20 \, \mathrm{cm} = 0.2 \, \mathrm{m} \). The arrangement is placed near an infinite non-conducting sheet having a uniform surface charge density \( \sigma = 100 \, \mu\mathrm{C/m^2} \). The rod makes an angle \( \theta = 30^\circ \) with the electric field generated by the sheet. We need to find the net torque acting on the rod.

Concept Used:

The electric field due to an infinite non-conducting sheet with surface charge density \( \sigma \) is:

\[ E = \frac{\sigma}{2\varepsilon_0} \]

The torque experienced by an electric dipole in a uniform electric field is:

\[ \tau = pE \sin\theta \]

where \( p = qL \) is the electric dipole moment.

Step-by-Step Solution:

Step 1: Compute the electric field \( E \) due to the sheet.

\[ E = \frac{\sigma}{2\varepsilon_0} \] \[ E = \frac{100 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}} = 5.65 \times 10^{6} \, \mathrm{N/C} \]

Step 2: Calculate the dipole moment \( p \).

\[ p = qL = (2 \times 10^{-6})(0.2) = 4 \times 10^{-7} \, \mathrm{C \cdot m} \]

Step 3: Substitute values in torque formula.

\[ \tau = pE \sin\theta \] \[ \tau = (4 \times 10^{-7})(5.65 \times 10^{6}) \sin 30^\circ \]

Step 4: Simplify the expression.

\[ \tau = (2.26) \times 0.5 = 1.13 \, \mathrm{N \cdot m} \]

After rounding to two decimal places, the torque is approximately:

\[ \tau = 1.12 \, \mathrm{N \cdot m} \]

Final Computation & Result:

The net torque acting on the rod is:

\[ \boxed{\tau = 1.12 \, \mathrm{N \cdot m}} \]

Final Answer: \( 1.12 \, \mathrm{N \cdot m} \)