Question

Two small objects each with a net charge of $ + Q $ exert a force of magnitude $ F $ on each other. We replace one of the objects with another whose net charge is $ + 4Q. $ We move the $ + Q $ and the $ + 4Q $ charges to be 3 times as far apart as they were. What is the magnitude of the force on the $ + 4Q $ charge?
(A) $ F $
(B) $ 4F $
(C) $ \dfrac{{4F}}{3} $
(D) $ \dfrac{{4F}}{9} $

Answer 1

Hint: We are provided with two conditions: find the force exerted on the charge $ + Q $ and on the charge $ + 4Q. $ if you substitute the force exerted on $ + Q $ in the value of force on the charge $ + 4Q $ then you will get the relation between them.

Complete step by step answer:
The force between two charges was studied by the scientist coulomb.
Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the square of the distance between them. The direction of this force is along the electric lines of force which are nothing but imaginary lines.
Let $ {q_1}{\text{ and }}{{\text{q}}_2} $ be the two point charges placed in air or vacuum at a distance r apart.
Then according to coulomb’s law the force of attraction between them is
 $ \Rightarrow F \propto \dfrac{{{q_1}{q_2}}}{{{r^2}}} $
 $ \Rightarrow F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}} $
K is the constant
 $ \Rightarrow k = \dfrac{1}{{4\pi {\varepsilon _0}}} $
 $ \Rightarrow F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} $
The unit of force is Newton.
The unit of charge is coulomb.
Given that, Initially there are two charges, each has a net charge of $ + Q $ separated by a distance r exert a force $ F $ on each other
The force exerted by the charges on each other by coulomb’s law is
 $ \Rightarrow F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Q \times Q}}{{{r^2}}} $
$ \Rightarrow F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q^2}}}{{{r^2}}}{\text{ }} \to {\text{1}} $
Then, one of the objects is replaced with another whose net charge is $ + 4Q. $
So, now there will be two charges $ + Q $ and $ + 4Q. $ which are three times far apart from each other. Then the distance between them is $ 3r. $ Let the force exerted on each other be $ F' $
The force exerted by the charges on each other by coulomb’s law is
 $ \Rightarrow F' = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{4Q \times Q}}{{{{(3r)}^2}}} $
 $ \Rightarrow F' = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{4{Q^2}}}{{9{r^2}}} $
 $ \Rightarrow F' = \dfrac{4}{9}\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q^2}}}{{{r^2}}} $
From equation 1 we get
 $ \Rightarrow F' = \dfrac{4}{9}F $
The magnitude of the force on the $ + 4Q $ charge is $ \dfrac{4}{9} $ times the force on $ + Q $
Hence the correct answer is option (D) $ \dfrac{4}{9}F. $

Note: Force is a vector quantity, it has both magnitude and direction. We have to consider both the magnitude and direction. If the direction of the force acting on the charges is opposite to each other they cancel out each other.