Question

$125$ drops are charged to a potential of $200V$ . These drops are combined to form a big drop. Calculate potential and charge in energy.
(A) $V = (\frac{1}{{2\pi \varepsilon }})\frac{q}{r} = (100)$
(B) $V = (\frac{1}{{4\pi \varepsilon }})\frac{q}{r} = (200)$
(C) $V = (\frac{1}{{6\pi \varepsilon }})\frac{q}{r} = (300)$
(D) $V = (\frac{1}{{8\pi \varepsilon }})\frac{q}{r} = (400)$

Answer 1

Hint: We will take in the concept that if a charge is equally distributed on a spherical body, then the body acts as if all the charge is concentrated in its center. Then we will find the radius of the big drop. Finally, we will calculate the potential.
Formulae Used: \[V = (\frac{1}{{4\pi \varepsilon }}){\kern 1pt} \frac{q}{r}\]
Where, $V$ is the potential, $q$ is the charge and $r$ is the radius of the charge.

Step By Step Solution
Let us assume that the small drops as well as the big drop are spherical.
Now,
Potential of each small drop, \[{V_s} = (\frac{1}{{4\pi \varepsilon }})\frac{q}{r}\]
$r$ is the radius of each of the small spherical drops.
Now,
\[Volume{\text{ }}of{\text{ }}the{\text{ }}Big{\text{ }}Spherical{\text{ }}drop = 125 \times Volume{\text{ }}of{\text{ }}each{\text{ }}small{\text{ }}drop\]
Now,
Volume of the big drop, \[Vo{l^m}_{big} = \frac{4}{3}\mathop {\pi R}\nolimits^3 \]
$R$ is the radius of the big drop.
Volume of each small drop, \[Vo{l^m}_{small} = \frac{4}{3}\mathop {\pi r}\nolimits^3 \]
Thus,
$\frac{4}{3}\mathop {\pi R}\nolimits^3 = 125 \times \frac{4}{3}\mathop {\pi r}\nolimits^3 $
Thus, we get
$R = 5r$
Now,
Net charge on the big drop, \[Q = 125 \times q\]
Again,
Potential of the big drop, \[\mathop V\nolimits_b = \frac{1}{{4\pi \varepsilon }}\frac{Q}{R} = \frac{1}{{4\pi \varepsilon }}\frac{{125q}}{{5r}} = 25 \times \mathop V\nolimits_s \]
Thus,
$V = \frac{1}{{4\pi \varepsilon }}\frac{q}{r} = (200)$

Thus, the answer is (B).

Additional Information: The term potential which we discussed now is analogic to work. It is actually a work done on bringing a charge say $\mathop q\nolimits_0 $ from infinity to a point of concern.
Thus its basic fundamental formula is given by
$V = \frac{W}{{\mathop q\nolimits_0 }}$
There is another term called the potential difference.
It is the work done on moving a charge $\mathop q\nolimits_0 $ from a point say $A$ to a point $B$ within the area of concern.
Thus,
The formula is given by
\[\Delta V = \frac{{\Delta W}}{{\mathop q\nolimits_0 }} = \frac{{\mathop W\nolimits_B - \mathop W\nolimits_A }}{{\mathop q\nolimits_0 }}\]

Note: We took \[Volume{\text{ }}of{\text{ }}the{\text{ }}Big{\text{ }}Spherical{\text{ }}drop = 125 \times Volume{\text{ }}of{\text{ }}each{\text{ }}small{\text{ }}drop\] because the total volume of the big drop is totally contributed by the accumulation of the small drops. This is the reason we took this calculation.