Question

Two electrons in a vacuum exert force of F=8.8 E-09 N on each other. They are then moved so that they are separated by x=6.4 times their original distance.

Part (a) What is the force, in newtons, that the electrons experience at the new separation distance?

Part (b) How far apart, in meters, were the electrons originally?

Answer 1

Part (b)

charge of an electron=1.60*10^-19 C

distance =?

$F=kq^{2}/d^{2}$

ANSWER:

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Part (a) Now distance is increased to 64 times

$F'=kq^{2}/d'^{2}$

$F'=9*10^{9}*(1.6*10^{-19})^{2}/(1.0368*10^{-9}m)^{2}$

ANSWER:

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