What is the repulsive electrical force between two protons in a
nucleus that are 4.6x 10-l5 m apart from each
other?
n.b. give your answer in Newtons
Electric force is given by the Coulomb's law
F = k* q1*q2/r2
where ,
K = 9* 109
q1 and q1 are charges in coulomb
r = distance between charges
charge of proton = 1.6* 10-19
r = 4.6*10-15
Therefore ,
F = 9*109 *1.6* 10-19 *1.6*10-19/(4.6*10-15)2
F= 1.08*101 N = 10.8 N
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