Q150 / 180 · COMEDK 2025 (25 May Shift) · Physics
A small metallic sphere carrying a charge $2 \times 10^{-6} C$ is suspended using a thin thread of negligible mass from a rigid support, in a horizontal electric filed $5 \times 10^4 \mathrm{Vm}^{-1}$. If the thread is making an angle $30^0$ with the vertical, What is the tension in the thread?
- A0.02 N
- Correct answerB0.2 N
- C2 N
- D20 N
Solution
Let $q = 2 \times 10^{-6} \text{ C}$ be the charge on the sphere, $E = 5 \times 10^4 \text{ Vm}^{-1}$ be the electric field, and $\theta = 30^{\circ}$ be the angle the thread makes with the vertical.
The forces acting on the sphere are the gravitational force $mg$ acting downwards, the electric force $F_e = qE$ acting horizontally, and the tension $T$ in the thread.
In equilibrium, the forces are balanced:
$T \sin \theta = qE$
$T \cos \theta = mg$
From the first equation, $T = \dfrac{qE}{\sin \theta}$.
Substituting the given values:
$T = \dfrac{(2 \times 10^{-6} \text{ C}) \times (5 \times 10^4 \text{ Vm}^{-1})}{\sin 30^{\circ}}$
$T = \dfrac{10^{-1}}{\dfrac{1}{2}}$
$T = 0.1 \times 2 = 0.2 \text{ N}$
Answer: 0.2 N
The forces acting on the sphere are the gravitational force $mg$ acting downwards, the electric force $F_e = qE$ acting horizontally, and the tension $T$ in the thread.
In equilibrium, the forces are balanced:
$T \sin \theta = qE$
$T \cos \theta = mg$
From the first equation, $T = \dfrac{qE}{\sin \theta}$.
Substituting the given values:
$T = \dfrac{(2 \times 10^{-6} \text{ C}) \times (5 \times 10^4 \text{ Vm}^{-1})}{\sin 30^{\circ}}$
$T = \dfrac{10^{-1}}{\dfrac{1}{2}}$
$T = 0.1 \times 2 = 0.2 \text{ N}$
Answer: 0.2 N
Hence, the correct answer is option: 'B'.