Question

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?

• A. 136.2 V
• B. 113 V
• C. 999.08 v
• D. 812.756 V

Answer 1

The correct option is D 812.756 V

$q=0.032nev=512Vv=k\frac{q}{r}\Rightarrow r=k\frac{q}{V}=\frac{{9.10}^9\times 0.032\times {10}^{-9}}{572}=5625\times {10}^{-4}m$

Volume(new)$=2\times volume$

$\frac{4}{3}\pi r(new{)}^3=2.\frac{4}{3}\pi r^{3}r(new)=\sqrt{2}rr(new)=3\sqrt{2}\times 5625\times {10}^{-4}=7.087\times {10}^{-4}(m)V=k\frac{q}{r}=9\times {10}^9\times \frac{2\times 0.032\times {10}^{-19}}{7.089\times {10}^{-4}}V=812.756V$