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.

• A. 660 V
• B. 1320 V
• C. 1520 V
• D. 1980 V

Answer 1

The Correct Option is D

To find the potential of the bigger drop formed by combining 27 smaller drops, each charged at \(220\, \text{V}\), we need to apply the principles of electrostatics and volume conservation.

1. When droplets combine, charge and volume are conserved. Given 27 drops each charged at \(220\, \text{V}\), let the charge on one drop be \(q\) and its radius be \(r\). Therefore, the total charge of the system is \(27q\).
2. The potential \(V\) of a spherical drop of radius \(r\) is given by: \(V = \frac{kq}{r}\) where \(k\) is Coulomb's constant.
3. Each small drop has a potential \(220\, \text{V}\): \(220 = \frac{kq}{r}\)
4. When 27 drops combine, volume conservation implies: \(\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3 \Rightarrow R = 3r\) where \(R\) is the radius of the bigger drop.
5. The potential of the bigger drop, \(V'\), is: \(V' = \frac{k \cdot 27q}{R} = \frac{k \cdot 27q}{3r} = 9 \cdot \left(\frac{kq}{r}\right)\)
6. Substituting the potential of the small drop: \(V' = 9 \times 220 = 1980\, \text{V}\)

Thus, the potential of the bigger drop is \(1980\, \text{V}\), which is the correct answer.