Question

Two-point charges A and B, having charges $ + Q$ and $ - Q$ respectively, are placed at certain distance apart and force acting between them is F. If $25\% $ charge of A is transferred to B, then force between the charges becomes
A. $F$
B. $\dfrac{{9F}}{{16}}$
C. $\dfrac{{16F}}{9}$
D. $\dfrac{{4F}}{3}$

Answer 1

Hint: The two-point charges are of equal and opposite magnitude. The force between them is based on Coulomb's law of electrostatic force between the charges. At first, we will calculate the force between the charges by applying Coulomb's law and then for the transferred charges we will calculate the value of force by using the given condition.

Complete step by step answer:
Coulomb’s law of electrostatic force states that the force of attraction or repulsion between the point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. The direction of force is along the line joining the two charges.
The formula is given as:
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here, $F$ is the force, $k$ is the proportionality constant, ${q_1},{q_2}$ are the charges,
$r$ is the distance between the charges.
We are given that the charges are $ + Q$ and $ - Q$ , therefore the force becomes
$F = k\dfrac{{Q( - Q)}}{{{r^2}}}$
$ \Rightarrow F = - k\dfrac{{{Q^2}}}{{{r^2}}}$
Taking magnitude only, we have
$ \Rightarrow F = k\dfrac{{{Q^2}}}{{{r^2}}}$ --equation $1$
After $25\% $ charge of A is transferred to B, the new charges are $Q - \dfrac{Q}{4}$ and $ - Q + \dfrac{Q}{4}$ respectively which is $\dfrac{{3Q}}{4}$ and $ - \dfrac{{3Q}}{4}$ .
Substituting these charges in the formula, we get
${F_0} = k\dfrac{{{{\left( {\dfrac{{3Q}}{4}} \right)}^2}}}{{{r^2}}}$
$ \Rightarrow {F_0} = k\dfrac{{9{{\left( Q \right)}^2}}}{{16{r^2}}}$ --equation $2$
From equation $1$ we know that
$F = k\dfrac{{{Q^2}}}{{{r^2}}}$
Substituting this value in equation $2$ we get
$\therefore{F_0} = \dfrac{{9F}}{{16}}$

Thus, option B is the correct option.

Note:k is the proportionality constant and the value of k is $$9 \times {10^9}\,N\,{m^2}\,{C^{ - 2}}$$ . k is known as Coulomb’s constant. The value of k totally depends on the medium in which the charged objects are placed. Please note, that it is easy to compare when the force is in terms of variables instead of substituting the value of Coulomb’s constant.