Question

If the force between the two charges is 60N, how the force if the distance is reduced by third and the two charges remain constant?

Answer 1

Solution.

$F_1=60N,$

$q_1=q_2=const,$

$r_2=r_1-\dfrac{r_1}{3};$

$F_1=k\sdot\dfrac{q_1\sdot q_2}{r_1^2};$

$F_2=k\sdot\dfrac{q_1\sdot q_2} {(\dfrac{2r}{3})^2};$

$F_2=\dfrac{9F_1}{4};$

$F_2=\dfrac{9\sdot60}{4}=135N;$

Answer: $F_2=135N.$