calculate the radius of a water drop which would just remain suspended in an electric field of 300 v/cm and charge with 1 electron.
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electric field (E) = kq/r^2 = 300 V/cm = 30000 v/m
q = ne
no. of electron = 1
q = 1 x (1.6 x 10^-19)
q = 1.6 x 10^-19
E = (9 x 10^9 x 1.6 x 10^-19)/r^2
r^2 = (9 x 16 x 10^-11)/30000
so r = 2.19 x 10^-7 m
i hope it will help you
regards
q = ne
no. of electron = 1
q = 1 x (1.6 x 10^-19)
q = 1.6 x 10^-19
E = (9 x 10^9 x 1.6 x 10^-19)/r^2
r^2 = (9 x 16 x 10^-11)/30000
so r = 2.19 x 10^-7 m
i hope it will help you
regards
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