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 charged body is placed in a horizontal electric field, it experiences two forces: the downward gravitational force ($mg$) and the horizontal electric force ($qE$). In equilibrium, the string makes an angle $\theta$ with the vertical.

Step 2: Identifying the Formula and Values:
In equilibrium, the horizontal and vertical components of tension $T$ are: $T \sin\theta = qE$ and $T \cos\theta = mg$ Dividing the two: $\tan\theta = \frac{qE}{mg}$ Given: $q = 50 \times 10^{-6}\text{ C}$, $E = 10 \times 10^3\text{ V/m}$, $m = 20 \times 10^{-3}\text{ kg}$, $g = 10\text{ m/s}^2$.

Step 3: Calculation:
\[ \tan\theta = \frac{(50 \times 10^{-6}) \times (10 \times 10^3)}{(20 \times 10^{-3}) \times 10} \] \[ \tan\theta = \frac{0.5}{0.2} = 2.5 \] \[ \theta = \tan^{-1}(2.50) \]

Step 4: Final Answer:
The angle is $\tan^{-1}(2.50)$.