Question 75: Electrostats & Capacitance

Question

Two point charges $$\mathrm{M}$$ and $$\mathrm{N}$$ having charges $$+q$$ and $$-q$$ respectively are placed at a distance apart. Force acting between them is $$\mathrm{F}$$. If $$30 \%$$ of charge of $$\mathrm{N}$$ is transferred to $$\mathrm{M}$$, then the force between the charges becomes:

Accepted Answer

To determine the potential of the bigger drop formed by the combination of 216 smaller drops, we need to consider the following principles of electrostatics and geometry:
1.
Volume Conservation:
The total volume of the bigger drop will be equal to the sum of the volumes of the 216 smaller drops. If each smaller drop has a radius $$r$$, then the volume of one small drop is:
$$V_{	ext{small}} = \dfrac{4}{3} \pi r^3$$
Since there are 216 such drops, the total volume becomes:
$$V_{	ext{total}} = 216 	imes \dfrac{4}{3} \pi r^3$$
Let $$R$$ be the radius of the bigger drop. The volume of the bigger drop is:
$$V_{	ext{big}} = \dfrac{4}{3} \pi R^3$$
By volume conservation, we have:
$$\dfrac{4}{3} \pi R^3 = 216 	imes \dfrac{4}{3} \pi r^3$$
or,
$$R^3 = 216 r^3$$
Therefore,
$$R = 6r$$
2.
Charge Conservation:
The total charge on the bigger drop will be the sum of the charges on all the smaller drops. If each smaller drop has a potential of $$200 \mathrm{~V}$$ and radius $$r$$, then its charge $$q$$ can be given by:
$$V_{	ext{small}} = \dfrac{k q}{r}$$
Given $$V_{	ext{small}} = 200 \mathrm{~V}$$, we get:
$$q = \dfrac{200 r}{k}$$
For 216 such drops, the total charge $$Q_{	ext{total}}$$ is:
$$Q_{	ext{total}} = 216 	imes \dfrac{200 r}{k} = \dfrac{43200 r}{k}$$
3.
Potential of the Bigger Drop:
The potential of the bigger drop can be calculated using its charge and radius. The formula for the potential of a spherical drop is:
$$V_{	ext{big}} = \dfrac{k Q_{	ext{total}}}{R}$$
Substitute the values:
$$V_{	ext{big}} = \dfrac{k \dfrac{43200 r}{k}}{6r} = \dfrac{43200 r}{6r} = 7200 \mathrm{~V}$$
Thus, the potential of the bigger drop is $$7200 \mathrm{~V}$$.
Answer:
Option B: 7200 V

Answer

F

Answer

$$
\frac{100}{49} F
$$

Answer

$$
\frac{49}{100} F
$$

Answer

$$
\frac{9}{16} F
$$

Electrostats & Capacitance

Two point charges $\mathrm{M} and \mathrm{N} having charges +q and -q respectively are placed at a distance apart. Force acting between them is \mathrm{F}. If 30 \% of charge of \mathrm{N} is transferred to \mathrm{M}$, then the force between the charges becomes:

Choose the correct answer: