A) 2 times
B) 4 times
C) 3 times
D) 8 times
Correct Answer: B
Solution :
Volume of bigger drop = 8 x volume of small drops \[\frac{4}{3}\pi {{R}^{3}}=8\times \frac{4}{3}\pi {{r}^{3}}\] \[R=2r\] Potential on small drop \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Potential on bigger drop \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{8q}{R}\] \[\begin{align} & =\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{8q}{2r} \\ & =4\left( \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r} \right)=4{{V}_{1}} \\ \end{align}\] So, potential on the bigger is four times to that potential on small.
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