Question

if an Oil Drop of a 3.2 *10 to the power minus13 N is balanced in an electric field of 5*10 to the power 5 v/m . find the charge on the oil drop?​

Answer 1

Answer:

F = E * Q     where E is the electric field and Q the charge

Q = F / E = 3.2 * 10E-13 / 5 * 10E5 = 6.4 * 10E-19 C

this will not give you the value of the elementary unit charge

which is 1.6 * 10E-19

Many measurements need to be made giving Q = n *e

and then determining the integral (integers) values for n that are consistent with the measurements because different drops will different numbers of charges on them