Question

One hundread twenty five small liquid drops each carrying a charge of 0.5 microcouloumb and each diameter of 0.1 m from a bigger drop . Calculate the potential at the surface of the bigger drop.

Answer 1

Given:

Total number of small liquid drops = 125

Diameter of small liquid drop (d) = 0.1 m

Charge on small drop = 0.5 μC

To find:

The potential at the surface of the bigger drop

Calculation:

As the smaller drops combine to form a bigger drop, the volume remains constant.

     $125\times\frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3$

     $125\times r^3=R^3$

     $R=5r$

     $R=5\times\frac{0.1}{2}$

     $R=0.25 m$

Potential of the drop is given by the formula

     $V=k\times\frac{Q}{R}$

     $V=k\times\frac{125\times q}{R}$

     $V= 9\times10^9\times\frac{125\times0.5\times10^{-6}}{0.25}$

     $V=225\times 10^4V$

Final answer:

The potential at the surface of bigger drop is 225×10⁴V