Question

Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of the bigger drop compared to each individual small drop is?
A. 8 times
B. 4 times
C. 2 times
D. 32 times

Answer 1

Hint: The capacitance of a sphere depends directly on its radius. When smaller drops combine to form a larger drop, then the radius of the larger drop can be calculated by considering the fact that the total volume of the smaller drops should be equal to the volume of the larger drop.
Formula used: Capacitance of a sphere of a radius $R$ is $4\pi \varepsilon R$ where $\varepsilon$ is the permittivity of the medium.
Volume of the sphere is given by $\dfrac{4}{3}\pi {{R}^{3}}$.

Complete step by step answer:

The capacitance of the body is a measure of the ability of the body to store electric charge when a potential difference is applied to the body.

The capacitance of a sphere of a radius $R$ is $4\pi \varepsilon R$--(1)

where $\varepsilon$ is the permittivity of the medium.

When many spheres combine to form a larger sphere, the total volume of the smaller spheres is equal to the volume of the larger sphere. Using this we will find a relation between radii of the smaller spheres and the larger sphere. Using this relation, we can find out the required relation between the individual capacitances of each small drop and the larger drop.
The volume of a sphere of radius $R$ is given by $\dfrac{4}{3}\pi {{R}^{3}}$. --(2)
Now, let us analyze the question.

We are given that eight drops of mercury of equal radii combine to form a larger drop.
Let the radius of the smaller drops be $r$.

Therefore, using (2), the volume of one small sphere will be,

$\dfrac{4}{3}\pi {{r}^{3}}$.

Therefore, total volume of 8 small spheres will be $8\times \dfrac{4}{3}\pi {{r}^{3}}$. --(3)
Let the capacitance of each individual small drop be ${{C}_{small}}$.

Using (1), the capacitance of each individual small drop would be ${{C}_{small}}=4\pi \varepsilon r$- (4).

Let the radius of the larger drop be $R$.

Therefore, using (2), the volume of one small sphere will be,
$\dfrac{4}{3}\pi {{R}^{3}}$. --(5)

Let the capacitance of each individual small drop be ${{C}_{l\arg e}}$.

Using (1), the capacitance of each individual small drop would be ${{C}_{l\arg e}}=4\pi \varepsilon R$- (6).

Now, since the smaller drops combine to form a larger drop, the total volume of the smaller drops should be equal to the volume of the larger drop. Therefore, using (3) and (5),

$8\times \dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi {{R}^{3}}$
$\therefore 8{{r}^{3}}={{R}^{3}}$
$\therefore {{\left( 2r \right)}^{3}}={{R}^{3}}$

Now putting cube root on both sides of the equation, we get,

$\sqrt[3]{{{\left( 2r \right)}^{3}}}=\sqrt[3]{{{R}^{3}}}$
$\therefore 2r=R$ ---(7)

Capacitance of bigger drop compared to the smaller drop is given by,

$\dfrac{{{C}_{l\arg e}}}{{{C}_{small}}}$
Using (4) and (6), in the ratio, we get,
$\dfrac{{{C}_{l\arg e}}}{{{C}_{small}}}=\dfrac{4\pi \varepsilon R}{4\pi \varepsilon r}=\dfrac{R}{r}$

Using equation (7) in the above equation,

$\dfrac{{{C}_{l\arg e}}}{{{C}_{small}}}=\dfrac{R}{r}=\dfrac{2r}{r}=2$
Hence, the capacitance of the large spherical drop is 2 times that of each individual small spherical drop.

Hence, the correct option is C) 2 times.

Note: Students might get confused as to how to proceed in the question since it is relating two very dissimilar topics, that is the capacitance and the volume of a spherical drop. However, they must realize that the common link joining these two is the radius of the drop. In such questions where, smaller objects join up to form a bigger object, one must always take this as a hint that they will have to equate either the total charge or the mass or the volume in the initial and final conditions since they remain conserved. Which quantity among these is to be equated has to be found out after analyzing and understanding the question.It is very important that students know about the capacitances of differently shaped bodies. Even if they do not know the exact formulas, they must know on which dimensional factors of the shape the capacitance depends. This helps a lot especially in these types of questions which in essence, are asking for a ratio in the answer.