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Answer 1 · 2017-01-21T09:07:34

Force becomes four times.

If #r# is the distance between two charges #q_1 and q_2# the force of interaction is given by Coulombs Law

#|vecF|=k_{e}\frac {q_{1}q_{2}}{r^{2}}# .....(1)
where #k_e# is Coulomb's constant# = 8.99×10^9 N m^2 C^-2#

Given that both charges are doubled. The new force is
#|vecF_"New"|=k_{e}\frac {(2q_{1})(2q_{2})}{r^{2}}#

#=>|vecF_"New"|=k_{e}\frac {4q_{1}q_{2}}{r^{2}}# ....(2)
Using equation (1) we have

#|vecF_"New"|=4|vecF|#