Q179 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
A tiny ball of mass $\mathrm{m}$ and charge $\mathrm{q}$ is suspended from the fixed support using an insulating string of length $1 \mathrm{~m}$. The horizontal uniform electric field $\mathrm{E}$ is switched on. The angle made by the string with vertical when the ball is in equilibrium is $45^{\circ}$. The magnitude of uniform electric field is
- A$E=\dfrac{m g}{\sqrt{2} q} \quad N C^{-1}$
- Correct answerB$E=\dfrac{m g}{q} \quad N C^{-1}$
- C$\mathrm{E}=0 \quad \mathrm{NC}^{-1}$
- D$\mathrm{E}=\dfrac{\sqrt{2} \mathrm{mg}}{\mathrm{q}} \mathrm{NC}^{-1}$
Solution
Let the mass of the ball be $m$, charge be $q$, and the electric field be $E$. The forces acting on the ball in equilibrium are:
1. Gravitational force $mg$ acting vertically downwards.
2. Electric force $qE$ acting horizontally.
3. Tension $T$ in the string acting along the string at an angle $\theta = 45^{\circ}$ with the vertical.
Resolving the forces into horizontal and vertical components:
Horizontal direction: $T \sin \theta = qE$
Vertical direction: $T \cos \theta = mg$
Dividing the two equations:
$\dfrac{T \sin \theta}{T \cos \theta} = \dfrac{qE}{mg}$
$\tan \theta = \dfrac{qE}{mg}$
Given $\theta = 45^{\circ}$, we have $\tan 45^{\circ} = 1$.
$1 = \dfrac{qE}{mg}$
$E = \dfrac{mg}{q}$
Answer: $E=\dfrac{m g}{q} \quad N C^{-1}$
1. Gravitational force $mg$ acting vertically downwards.
2. Electric force $qE$ acting horizontally.
3. Tension $T$ in the string acting along the string at an angle $\theta = 45^{\circ}$ with the vertical.
Resolving the forces into horizontal and vertical components:
Horizontal direction: $T \sin \theta = qE$
Vertical direction: $T \cos \theta = mg$
Dividing the two equations:
$\dfrac{T \sin \theta}{T \cos \theta} = \dfrac{qE}{mg}$
$\tan \theta = \dfrac{qE}{mg}$
Given $\theta = 45^{\circ}$, we have $\tan 45^{\circ} = 1$.
$1 = \dfrac{qE}{mg}$
$E = \dfrac{mg}{q}$
Answer: $E=\dfrac{m g}{q} \quad N C^{-1}$
Hence, the correct answer is option: 'B'.