Question 79: Electrostats & Capacitance

Question

Three point charges are located on a circular arc at $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ as shown in the figure below. The total electric field at the centre of the $$\operatorname{arc}(\mathrm{C})$$ 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

$$15000 \mathrm{~NC}^{-1}$$

Answer

$$10000 \mathrm{~N} \mathrm{C}^{-1}$$

Answer

$$20000 \mathrm{~NC}^{-1}$$

Answer

$$5000 \mathrm{~NC}^{-1}$$

Electrostats & Capacitance

Three point charges are located on a circular arc at $\mathrm{A}, \mathrm{B} and \mathrm{C} as shown in the figure below. The total electric field at the centre of the \operatorname{arc}(\mathrm{C})$ is

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