Question

N spherical droplets each of radius r, have been charged to have a potential of V each. If all these drops have to coalesce to form a single large droplet , what would be the potential of this large dropï¼

Answer 1

Answer:

Hope it is helpful to you

Answer 2

$V \times {N}^{ \frac{2}{3}}$

Explanation:

$we \: \: know \: \: that \: \: \boxed{ potential\: (V)= \frac{1}{4\pi \epsilon_{o} } \frac{q}{r} } \\ Let, \: \: \: radius \: \: of \: \: a \: \: big \: \: drop \: \: formed \: \: by \: \: joining \: \: N\: \: drops = r_{b} \\Now , \: \: volume \: \: of \: \: large \: \: drop = N \times volume \: \: of \: \: small \: \: drops \\ \frac{4}{3} \pi {r_{b} }^{3} = N \times \frac{4}{3} \pi {r }^{3} \\ {r_{b} }^{3} = N \times {r}^{3} \\ r_{b} = N^{ \frac{1}{3} } \times {r} \\A/Q, \: \: large \: \: drop \: \: potential = \frac{1}{4\pi \epsilon_{o} } \frac{N \times q}{r_{b} } \\ putting \: \: value \: \: of \: \: r_{b} \\ then, \: \: large \: \: drop \: \: potential= \frac{1}{4\pi \epsilon_{o} } \frac{N\times q}{ {N}^{ \frac{1}{3} } \times r}\\ =\frac{1}{4\pi \epsilon_{o} } \frac{q}{ r} \times {N}^{ \frac{2}{3}} \\ = V \times {N}^{ \frac{2}{3}}$