Sixty four drops of radius `0.02` m and each carrying a charge of `5muC` are combined to form a bigger drop. Find, how the surface charge density of electrification will change, if no charge is lost.

Explanation

Volume of each small drop `=(4)/(3)pi(0.02)^(3)m^(3)`
Volume of 64 snall drops `=(4)/(3)pi(0.02)^(3)xx64m^(3)`
Let R be the radius of the bigger drop formed. Then,
`(4)/(3)piR^(3)=(4)/(3)pi(0.02)^(3)xx64` or `R^(3)=(0.02)^(3)xx4^(3)`
`:." "R=0.02xx4=0.08` m
Charge on small drop, `q=5muC=5xx10^(-6)C`
`:.` Surface charge density of small drop,
`sigma_(1)=(q)/(4pir^(2))=(5xx10^(-6))/(4pi(0.02)^(2))"Cm"^(-1)`
`:.` Surface charge density of bigger drop,
`sigma_(2)=(5xx10^(-6)xx64)/(4pi(0.08)^(2))"Cm"^(-2)`
`:.(sigma_(1))/(sigma_(2))=(5xx10^(-6))/(4pi(0.02)^(2))xx(4pi(0.08)^(2))/(5xx10^(-6)xx64)=1/4`