Question

Two charged particles are 8.45cm apart.They are moved and the force on each of them is found to have tripled.How far are they now?

Answer 1

\[F = \frac{k q _{1}q _{2} }{ .0845^2 }\]
\[3F = \frac{k q _{1}q _{2} }{ r^2 }\]
\[\frac{ 3F }{F } = \frac{k q _{1}q _{2} .0845^2 }{k q _{1}q _{2} r^2 }\]
\[3= \frac{.0845^2 }{r^2 }\]
\[r= \sqrt{\frac{.0845^2 }{3}}\]