Question

Two charged objects originally felt 12N of attraction. One charge is changed from to 3C to 6C and their distance changes from 15cm apart to 45cm apart. What is the new force of attraction ?

Answer 1

Given that,

Initial force, F = 12 N

First initial charge, q₁ = 3C

First new charge, q₁' = 6C

Initial distance, r = 15 cm

New distance, r' = 45 cm

To find,

The new force of attraction.

Solution,

The force between two charges is given by :

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

$12=k\dfrac{3\times q_2}{15^2}\ ....(1)$

Let F' is the new force.

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

$F'=\dfrac{k\times 6\times q_2}{(45)^2}\ ...(2)$

As q₂ is same in this case.

Dividing equation (1) and (2) :

$\dfrac{F}{F'}=\dfrac{k\dfrac{3q_2}{15^2}}{\dfrac{k\times 6\times q_2}{45^2}}\\\\\dfrac{12}{F'}=4.5\\\\F'=\dfrac{12}{4.5}\\\\F'=2.67\ N$

So, the new force of attraction is 2.67 N.