Question 81: Electrostatics & 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}
$

Electrostatics & Capacitance

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