64 drops of radius 0 02m having 5 microcoulomb charge each, combine to form a bigger drop How - Physics - Electric Charges And Fields

Question

64 drops of radius 0 02m having 5 microcoulomb charge each, combine
to form a bigger drop How surface density of electrification will
change if no charge is lost ?

Jyoti Pant · Accepted Answer

Surface area of each drop=4πr2=4π(0
02)2=16π×10-4m2Surface charge density of each drop, σ=q4πr2=5×10-616π×10-4=1320πCm-2Now, volume of a small drop=43πr3Volume of 64 drops=43πr3×64=43π(4r)3m3i
e
 volume of bigger drop is such that it has a radius of 4r and its total charge is 64×5×10-6=320×10-6CThus, surface charge density=64×5×10-64π×(4r)2Now,Surface charge density of smaller dropSurface charge density of bigger drop=1320π×4π(4πr)264×5×10-6=180×(4×0
02)264×5×10-6=14Thus, required ratio=1:4

64 drops of radius 0.02m having 5 microcoulomb charge each, combine to form a bigger drop . How surface density of electrification will change if no charge is lost ?