Question

2Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 1 is one-fourth the original value AND the charge of object 2 is tripled AND the distance separating Objects 1 and 2 is halved, then the new electrostatic force will be _____ units.

Answer 1

Answer:

The new force becomes 3 times of the initial force.

Explanation:

Let q₁ and q₂ are two charged particles. The force between them is given by :

$F=\dfrac{kq_1q_2}{r^2}\ ......(1)$

If $q_1'=\dfrac{q_1}{4}$

and

$q_2'=3q_2$

Also, r' = r/2

New force,

$F'=\dfrac{kq_1'q_2'}{r'^2}$

Putting all the above values,

$F'=\dfrac{k\dfrac{q_1}{4}3q_2}{(r/2)^2}\\\\F'=3\times \dfrac{kq_1q_2}{r^2}\\\\F'=3F$

So, the new force becomes 3 times of the initial force.