Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

Concept:

The amount of work done to move a charge from one point to another point against the electric field is calculated using the formula:

\(V=\frac{kq}{r}\)

Where V is electric potential, k is coulomb's constant, q is charge, r is distance between two points.

Calculation:

Given-

n = 27

Q : Total charge on big drop

R : Radius of big drop

Let charge and radius of smaller drop is q and r respectively

For smaller drop, 

Let R be radius of bigger drop,

Volume = \(\frac{4}{3}\pi R^3 \)

As volume remains the same

\(\frac{4}{3}\pi R^3 = 27\times\frac{4}{3}\pi R^3\)

 ⇒ R = 3r

The potential of the small drop

\(\frac{KQ}{r}=v \)

The potential of the big drop

\(V^\prime=\frac{KQ}{R} \)            ---  (1)

Now, using charge conservation,

Q = n q

where n is number of electrons

Putting the value of Q in the equation (1),

\(V^\prime=\frac{K\times27q}{3r}=9v\)

The potential of bigger drop = 9 × 220 = 1980 V