Question

729 identical mercury drops with each one having a charge 1 pC and a radius 9 mm. If they are mixed and forming a big drop then find out potential over the big drop.

Answer 1

The charge on the big drop is simply the sum of all the individual charges of the small drops. Investigate the relationship of the radius of the big drop relative to the small drop noting that the volume of the sum of the small drop is the volume of the big drop.

Formula used: In this solution we will be using the following formulae;
$V = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{r}$ where $V$ is the electric potential of a charge at a particular point, ${\varepsilon _o}$ is the permittivity of free space, $q$ is the charge, and $r$ is the distance of the charge from the point of interest.
$Vol = \dfrac{4}{3}\pi {r^3}$ where $Vol$ is the volume of a spherical object, and $r$ is the radius of the object.

Complete Step-by-Step Solution:
To solve, let’s let the individual small drops of mercury have a charge $q$. Then,
$q = 1pC = 1 \times {10^{ - 12}}C$.
If the radius is $r$, then
$r = 9mm = 9 \times {10^{ - 3}}m$
For the big drop, the charge would be equal to the sum of the entire 729 small drops, hence
$Q = 729 \times 1 \times {10^{ - 12}}C = 7.29 \times {10^{ - 10}}C$
As for the radius, we have,
We note that the volume of the big drop is equal to the total volume of the small drops combined, hence
$\dfrac{4}{3}\pi {R^3} = 729\left( {\dfrac{4}{3}\pi {r^3}} \right)$ since $Vol = \dfrac{4}{3}\pi {r^3}$ where $Vol$ is the volume of a spherical object, and $r$ is the radius of the object.
Hence, by cancellation of common terms, we have
${R^3} = 729 \times {r^3}$
$\Rightarrow R = \sqrt[3]{{729{r^3}}} = 9r = 8.1 \times {10^{ - 2}}m$
Now electric potential can be given as
$V = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{q}{r}$ where ${\varepsilon _o}$ is the permittivity of free space, $q$ is the charge, and $r$ is the distance of the charge from the point of interest.
Then the potential over the big drop would be
$V = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{7.29 \times {{10}^{ - 10}}C}}{{8.1 \times {{10}^{ - 2}}m}}$
By computation knowing that $\dfrac{1}{{4\pi {\varepsilon _o}}} = 9.0 \times {10^9}N{m^2}/{C^2}$, we have
$V = 81{\text{ V}}$

Note: For clarity, note that this position would not be an equilibrium position. This is because the many charges around the big drop will repel each other outwards and hence will take the surface of the mercury along increasing the volume until the surface tension of the mercury balances the force.