Question

One charge is decreased to one-third of its original value,

Answer 1

Answer:

the Third option

Explanation:

Answer 2

Answer:

It will decrease to one-sixth the original force.

Explanation:

The electrical force is given by :

$F=k\dfrac{q_1q_2}{r^2}$..............(1)

Where

k is the electrostatic constant

r is the distance between charges

One charge is decreased to one-third of its original value,  and a second charge is decreased to one-half of its  original value.

i.e. $q_1'=\dfrac{q_1}{3}$

and $q_2'=\dfrac{q_2}{2}$

New force is given by :

$F'=k\dfrac{q_1q_2}{6r^2}$

$F'=\dfrac{F}{6}$

So, the new force will decrease to one-sixth the original force. Hence, the correct option is (c).