Question

$64$ identical mercury ball (each of potential $10 V$) combine to form a big ball, then the potential at the surface of the big ball will be:

Answer 1

Hint: Potential on the small drop is given. $64$ drops of mercury ball combine to form a big ball of radius R. We first find the relation between the radius of small drop and big drop. Charge on the big sphere is equal to the sum of the charges on the smaller sphere. Then we take the ratio of potentials of both sized spheres.

Complete step-by-step solution:
Capacitor of a sphere, $C = 4 \pi \epsilon_{o} r$
Potential is given by: $V = \dfrac{Q}{C}$
Let q be the charge on a small drop.
Q be the charge on a big charge.
$Q = 64q$
When $64$ drops of radius r forms a single drop of radius R.
$\therefore 64 \times \dfrac{4}{3} \pi r^{3} =\dfrac{4}{3} \pi R^{3} $
$R = 4r$
Potential on the small drop,
$V_{s} = \dfrac{q}{4 \pi \epsilon_{o} r }$
Potential on the big drop,
$V_{b} = \dfrac{64q}{4 \pi \epsilon_{o} R }$
Divide potential on big drops with the potential on small drops.
$\dfrac{V_{b}}{V_{s}} = \dfrac{64}{R} \times \dfrac{r}{1}$
Put $R = 4r$ in the above formula.
$\dfrac{V_{b}}{V_{s}} = \dfrac{64}{4r} \times \dfrac{r}{1}$
$V_{b} = 16 V_{s}$
Given: Potential on small drop, $ V_{s} = 10 V$
$V_{b} = 16 \times 10 = 160V$
The potential at the surface of the big ball will be $160V$.

Note: The electric potential is the measure of work energy required to run a unit of electrical charge from an origin point to a specific point in an energetic field with the negligible acceleration of the charge to avoid generating kinetic energy or radiation by charge. Typically, the origin point is the Earth at infinity, although any point can be utilized.