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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

$|vecF_"New"|=4|vecF|$