Question:

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.

Updated On: Jul 18, 2024
  • 660 V
  • 1320 V
  • 1520 V
  • 1980 V
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The Correct Option is D

Solution and Explanation

If each drop has a charge ' $q$ ' and radius ' $r$ '. Then from conservation of charge, charge on the big drop is $n q=27 q(n=27)$ from conservation of volume $\frac{4}{3} \pi r^{3} n=\frac{4}{3} \pi R^{3}$ $R=n^{1 / 3} r$ Now potential of the small drop $V=\frac{q}{4 \pi \epsilon_{0} r}=220 V$ Potential of the big drop, $V=\frac{n q}{4 \pi \in_{0} R}=\frac{n q}{4 \pi \in_{0} n^{1 / 3} r}=n^{2 / 3} \frac{q}{4 \pi \in_{0} r} $ $V=(27)^{2 / 3} \times 220 V$ $=9 \times 220=1980 V$
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Concepts Used:

Electrostatic Potential

The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as “The amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.”

SI Unit of Electrostatic Potential:

SI unit of electrostatic potential - volt

Other units - statvolt

Symbol of electrostatic potential - V or φ

Dimensional formula - ML2T3I-1

Electric Potential Formula:

The electric potential energy of the system is given by the following formula:

U = 1/(4πεº) × [q1q2/d]

Where q1 and q2 are the two charges that are separated by the distance d.