Multiple choice

A spherical drop of water carrying a charge of 0.032 nC has a potential of 512 V at its surface. If two such drops with the same radius and charge were to combine to form a single drop, what would be the potential at the surface of the new drop?

  1. 136.2 V

  2. 113 V

  3. 999.08 v

  4. 812.756 V
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D Correct answer
Explanation

Potential V = kQ/R. When two drops combine, volume doubles, so R_new = R * 2^(1/3). Charge Q_new = 2Q. V_new = k(2Q) / (R * 2^(1/3)) = V * 2^(2/3) = 512 * 1.5874 = 812.756 V.

AI explanation

The potential at the surface of a sphere is given by V = kQ/r. For the original drop, 512 = k(0.032 nC)/r, so the radius r is proportional to the charge. When two identical drops combine, the new charge is 2Q and the new volume is twice the original volume, making the new radius the cube root of 2 times r. The new potential V = k(2Q)/(r * cube root of 2). By dividing this by the original potential, V/512 = 2/(cube root of 2), which simplifies to V = 512 * cube root of 4. This yields a new potential of approximately 812.756 V.

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