Question

A body of $20\text{ g}$ and $+50\ \mu\text{C}$ charge is suspended from the ceiling by a string. If a horizontal electric field of $10\text{ kV/m}$ is applied, the angle between the string and vertical direction is: (Take $g = 10\text{ N/kg}$)

• A. $\tan^{-1}(2.50)$
• B. $\tan^{-1}(3.00)$
• C. $\tan^{-1}(3.50)$
• D. $\sin^{-1}(2.50)$

Hint:

For any particle in a horizontal field, the angle $\theta$ is always determined by the ratio of the horizontal force to the vertical force: $\tan\theta = F_h / F_v$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When the suspended body is in equilibrium, three forces act on it: the tension in the string (\( T \)), the weight (\( mg \)) acting vertically downwards, and the electric force (\( qE \)) acting horizontally.
Step 2: Key Formula or Approach:
For equilibrium at an angle \( \theta \) with the vertical:
Horizontal components: \( T \sin \theta = qE \)
Vertical components: \( T \cos \theta = mg \)
Dividing these gives: \( \tan \theta = \frac{qE}{mg} \)
Step 3: Detailed Explanation:
Given:
\( m = 20 \, \text{g} = 0.02 \, \text{kg} \)
\( q = 50 \, \mu\text{C} = 50 \times 10^{-6} \, \text{C} \)
\( E = 10 \, \text{kV/m} = 10^4 \, \text{V/m} \)
\( g = 10 \, \text{m/s}^2 \)
Calculation:
\[ qE = (50 \times 10^{-6}) \times (10^4) = 0.5 \, \text{N} \]
\[ mg = 0.02 \times 10 = 0.2 \, \text{N} \]
\[ \tan \theta = \frac{0.5}{0.2} = 2.5 \]
\[ \theta = \tan^{-1}(2.5) \]
Step 4: Final Answer:
The angle between the string and the vertical direction is $\tan^{-1}(2.50)$.