Question

Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them?a) 3F/2b) F/3c) F/6d) F/12e) 3F/4

Answer 1

Answer:

New force, $F'=\dfrac{F}{12}$

Explanation:

Given that, two point charges attract each other with an electric force of magnitude F. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

If one charge is reduced to one-third its original value and the distance between the charges is doubled such that,

$q_1'=\dfrac{q_1}{3}$, $r'=2r$

$F'=k\dfrac{q_1'q_2'}{r'^2}$

$F'=k\dfrac{(q_1/3)q_2}{(2r)^2}$

$F'=\dfrac{F}{12}$

So, the electric force between them is reduced to (1/12). Hence, the correct option is (d).