In the nucleus of an atom, two protons are separated by a distance of 1 x 10 -5m. What is the magnitude of the electric force between them?
Electrostatic force is given by:
F = k*q1*q2/r^2
k = 9*10^9
q1 = q2 = charge on proton = +1.6*10^-19 C
r = distance between two protons = 1*10^-5 m
So Using these values:
F = 9*10^9*(1.6*10^-19)^2/(1*10^-5)^2
F = 2.30*10^-18 N
Since both charge are positive, So force between them will be repulsive.
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