Question 81: Electrostats & Capacitance

Question

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

Accepted Answer

Let's analyze the forces acting on the ball when it is in equilibrium:
The ball of mass $$m$$ experiences a gravitational force downward: $$mg$$.
The ball experiences a horizontal electric force due to the uniform electric field, which is given by: $$qE$$.
The string exerts a tension $$T$$ along its length, making an angle of $$45^\circ$$ with the vertical.
Since the ball is in equilibrium, the forces in both the vertical and horizontal directions must balance:
Vertical component:
The vertical component of the tension is
$$T \cos 45^\circ$$
This balances the weight:
$$T \cos 45^\circ = mg.$$
Horizontal component:
The horizontal component of the tension is
$$T \sin 45^\circ$$
This balances the electric force:
$$T \sin 45^\circ = qE.$$
Now, divide the horizontal equation by the vertical equation to eliminate $$T$$:
$$ \frac{T \sin 45^\circ}{T \cos 45^\circ} = \frac{qE}{mg}. $$
Since
$$ \frac{\sin 45^\circ}{\cos 45^\circ} = 	an 45^\circ = 1, $$
the equation simplifies to:
$$ 1 = \frac{qE}{mg}. $$
Solving for $$E$$ gives:
$$ E = \frac{mg}{q}. $$
Thus, the magnitude of the uniform electric field is
$$ E = \frac{mg}{q} \quad 	ext{NC}^{-1}. $$
This corresponds to Option A.

Answer

$$
E=\frac{m g}{q} \quad N C^{-1}
$$

Answer

$$
\mathrm{E}=\frac{\sqrt{2} \mathrm{mg}}{\mathrm{q}} \mathrm{NC}^{-1}
$$

Answer

$$
E=\frac{m g}{\sqrt{2} q} \quad N C^{-1}
$$

Answer

$$
\mathrm{E}=0 \quad \mathrm{NC}^{-1}
$$

Electrostats & Capacitance

Choose the correct answer: