A simple pendulum with a bob (mass $ m $ and charge $ q $) is in equilibrium in the presence of a horizontal electric field $ E $. Then, the tension in the thread is $ F_1 $. Given $ \frac{F_1}{F_2} = \frac{2}{\sqrt{\alpha}} $, find $ \alpha $.

Question

A point charge $q = 1\,\mu\text{C}$ is located at a distance $2\,\text{cm}$ from one end of a thin insulating wire of length $10\,\text{cm}$ having a charge $Q = 24\,\mu\text{C}$, distributed uniformly along its length, as shown in the figure. Force between $q$ and wire is ________ N.

Answer 1

To solve this problem, we need to analyze the forces acting on the pendulum bob, which has mass $ m $ and charge $ q $, in the presence of an electric field $ E $.

Step-by-Step Solution:

Identify the forces acting on the bob:
- Gravitational Force: $ F_g = mg $ acting downward.
- Electric Force: $ F_e = qE $ acting horizontally to the right.
- Tension in the thread: $ T $ acting along the thread.
In equilibrium, the net force on the bob is zero. Therefore, we can resolve the tension $ T $ into its components:
- Vertical component: $ T \cos \theta = mg $
- Horizontal component: $ T \sin \theta = qE $
Divide the horizontal component equation by the vertical component equation to eliminate $ T $: $$ \frac{T \sin \theta}{T \cos \theta} = \frac{qE}{mg} $$ $$ \tan \theta = \frac{qE}{mg} $$
The tension in the string $ T $ when the pendulum is in equilibrium can be determined by using: $$ T = \sqrt{(mg)^2 + (qE)^2} $$
Given that $\frac{F_1}{F_2} = \frac{2}{\sqrt{\alpha}}$ and since $ F_1 = T $, the expression simplifies to: $$ \frac{T}{T_0} = \frac{2}{\sqrt{\alpha}} $$ where $ T_0 = mg $.
Substitute the expression for $ T $ to find: $$ \frac{\sqrt{(mg)^2 + (qE)^2}}{mg} = \frac{2}{\sqrt{\alpha}} $$
Squaring both sides, we get: $$ \frac{(mg)^2 + (qE)^2}{(mg)^2} = \frac{4}{\alpha} $$ $$ 1 + \left(\frac{qE}{mg}\right)^2 = \frac{4}{\alpha} $$
Rearranging gives: $$ \alpha = \frac{4}{1 + \left(\frac{qE}{mg}\right)^2} $$
From the options, by checking and substituting, we find that: $$ \alpha = 3 $$

Hence, the correct answer is $ \alpha = 3 $.

Answer 2

To solve this problem, we need to analyze the forces acting on the pendulum bob, which has mass $ m $ and charge $ q $, in the presence of an electric field $ E $.

Step-by-Step Solution:

Identify the forces acting on the bob:
- Gravitational Force: $ F_g = mg $ acting downward.
- Electric Force: $ F_e = qE $ acting horizontally to the right.
- Tension in the thread: $ T $ acting along the thread.
In equilibrium, the net force on the bob is zero. Therefore, we can resolve the tension $ T $ into its components:
- Vertical component: $ T \cos \theta = mg $
- Horizontal component: $ T \sin \theta = qE $
Divide the horizontal component equation by the vertical component equation to eliminate $ T $: $$ \frac{T \sin \theta}{T \cos \theta} = \frac{qE}{mg} $$ $$ \tan \theta = \frac{qE}{mg} $$
The tension in the string $ T $ when the pendulum is in equilibrium can be determined by using: $$ T = \sqrt{(mg)^2 + (qE)^2} $$
Given that $\frac{F_1}{F_2} = \frac{2}{\sqrt{\alpha}}$ and since $ F_1 = T $, the expression simplifies to: $$ \frac{T}{T_0} = \frac{2}{\sqrt{\alpha}} $$ where $ T_0 = mg $.
Substitute the expression for $ T $ to find: $$ \frac{\sqrt{(mg)^2 + (qE)^2}}{mg} = \frac{2}{\sqrt{\alpha}} $$
Squaring both sides, we get: $$ \frac{(mg)^2 + (qE)^2}{(mg)^2} = \frac{4}{\alpha} $$ $$ 1 + \left(\frac{qE}{mg}\right)^2 = \frac{4}{\alpha} $$
Rearranging gives: $$ \alpha = \frac{4}{1 + \left(\frac{qE}{mg}\right)^2} $$
From the options, by checking and substituting, we find that: $$ \alpha = 3 $$

Hence, the correct answer is $ \alpha = 3 $.

A simple pendulum with a bob (mass \( m \) and charge \( q \)) is in equilibrium in the presence of a horizontal electric field \( E \). Then, the tension in the thread is \( F_1 \). Given \( \frac{F_1}{F_2} = \frac{2}{\sqrt{\alpha}} \), find \( \alpha \).

Show Hint

The Correct Option is B

Solution and Explanation

Step-by-Step Solution:

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