Question

Two point charges of +5 uC are so
placed that they experience a force of
8.010-3N. They are then moved apart,
so that the force is now 2.010-3N. The
distance between them is now
(A) 1/4 the previous distance
(B) double the previous distance
(C) four times the previous distance
(D) half the previous distance​

Answer 1

Double the previous distance

Explanation:

Charges, $q_1=q_2=5\ \mu C=5\times 10^{-6}\ C$

Force, $F=8\times 10^{-3}\ N$

New force, $F'=2\times 10^{-3}\ N$

The electric force is given by :

$F=\dfrac{kq^2}{r^2}$

r is the initial distance

$8\times 10^{-3}=\dfrac{kq^2}{r^2}$................(1)

New electric force is given by :

$F'=\dfrac{kq^2}{r'^2}$

r' is the final distance

$2\times 10^{-3}=\dfrac{kq^2}{r'^2}$...........(2)

Dividing equation (1) and (2) :

$4=\dfrac{r'^2}{r}$

$r'=2r$

So, the distance between them is now double the previous distance. Hence, the correct option is (b).

Learn more :

Topic : Electrostatic force

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