Question

A spherical drop of capacitance \[12\mu F\] is broken into eight drops of equal radius. What is the capacitance of each small drop in \[\mu F\]?
A. 6
B. 4
C. 3
D. 1

Answer 1

Hint: In this question, we need to determine the capacitance of each drop-in micro-farads such that a spherical drop of \[12\mu F\] is broken. For this, we will use the relation between voltage, charge and the radius of the sphere.

Complete step by step answer:
Electrical potential on a charged sphere is given by the formula
\[V = \dfrac{{kQ}}{R}\], where \[Q\] is the charge on the sphere, \[R\] is the radius of that sphere and \[k\] is the constant which is equal to \[9 \times {10^9}N\].
The capacitance of spherical drop \[C = 12\mu F\] .We know that the electrical potential on a charged sphere is given by the formula
\[V = \dfrac{{kQ}}{R} - - (i)\]
The charge on a capacitor plate is determined by the formula
\[Q = CV - - (ii)\]
Now we substitute the equation (ii) in equation (i) to find the capacitance on the sphere; hence we get
\[
  V = \dfrac{{k\left( {CV} \right)}}{R} \\
  1 = \dfrac{{kC}}{R} \\
  C = \dfrac{R}{k} - - (iii) \\
 \]
Since the capacitance on the spherical drop is given as \[C = 12\mu F\] hence, we can write equation (iii) as
\[\dfrac{R}{k} = 12\mu F\]
Now it is said that spherical drop is broken into eight drops of equal radius, let us assume that the radius of the small drop is\[r\], so by comparing the volume of the smaller drops and the larger drop we can write
\[8\left( {\dfrac{4}{3}\pi {r^3}} \right) = \dfrac{4}{3}\pi {R^3}\]
Hence by solving we get
\[
  8\left( {\dfrac{4}{3}\pi {r^3}} \right) = \dfrac{4}{3}\pi {R^3} \\
\Rightarrow {\left( {\dfrac{R}{r}} \right)^3} = 8 \\
 \therefore \dfrac{R}{r} = 2 - - (iv) \\
 \]
So the capacitance on each smaller drop will be equal to
\[V = \dfrac{r}{k} - - (iv)\]
Now we substitute the value of r from equation (iv), we get
\[V = \dfrac{R}{{2k}}\]
Where \[\dfrac{R}{k} = 12\mu F\], hence the value of capacitance on smaller drop will be
\[
  V = \dfrac{{12}}{2} \\
   = 6\mu F \\
 \]

So, the correct answer is Option A.

Note:
Students must note that the electric potential on a charged sphere reduces when that charged sphere is divided into smaller spheres.
It is said that a spherical drop is broken into eight drops of equal radius, let us assume that the radius of the small drop.