Question

Two particles having charges ${{\text{q}}_1}$ and ${{\text{q}}_2}$ exert a force F on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled?
${\text{A}}{\text{.}}$ 12F
${\text{B}}{\text{.}}$ 4F
${\text{C}}{\text{.}}$ 8F
${\text{D}}{\text{.}}$ 16F

Answer 1

Hint: Here, we will proceed by expressing the force F in terms of the given charges and the separation distance (assumed). Then, we will make another relation between the new force when the distance between them is reduced to half and the charge on each particle is doubled.

Complete step-by-step answer:

Formula Used- ${\text{F}} = \dfrac{{{\text{K}}{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}$.
Given, Initial charge on the first particle = ${{\text{q}}_1}$
Initial charge on the second particle = ${{\text{q}}_2}$

Let the initial distance between two given stationary charges ${{\text{q}}_1}$ and ${{\text{q}}_2}$ be r

i.e., Initial distance of separation = r
As we know that according to Coulomb’s inverse square law, the electric force acting between two stationary charges ${{\text{q}}_1}$ and ${{\text{q}}_2}$ when separated by a distance of r is given by

Initial electric force experienced, ${\text{F}} = \dfrac{{{\text{K}}{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}{\text{ }} \to {\text{(1)}}$ where K denotes the Coulomb’s constant

According to the problem statement,

Charge on each particle is doubled

i.e., Final charge on the first particle = 2(Initial charge on the first particle) = 2${{\text{q}}_1}$
Final charge on the second particle = 2(Initial charge on the second particle) = 2${{\text{q}}_2}$
Also, the distance of the charges is reduced to half

i.e., Final distance of separation = $\dfrac{{\text{r}}}{2}$

Let ${{\text{F}}_1}$ be the final electric force acting between 2${{\text{q}}_1}$ and 2${{\text{q}}_2}$ which are separated by a distance of $\dfrac{{\text{r}}}{2}$

Now, replacing ${{\text{q}}_1}$ with 2${{\text{q}}_1}$, ${{\text{q}}_2}$ with 2${{\text{q}}_2}$, r with $\dfrac{{\text{r}}}{2}$ and F with ${{\text{F}}_1}$ in equation (1), we get
$
  {{\text{F}}_1} = \dfrac{{{\text{K}}\left( {{\text{2}}{{\text{q}}_1}} \right)\left( {{\text{2}}{{\text{q}}_2}} \right)}}{{{{\left( {\dfrac{{\text{r}}}{2}} \right)}^2}}} \\
   \Rightarrow {{\text{F}}_1} = \dfrac{{{\text{4K}}{{\text{q}}_1}{{\text{q}}_2}}}{{\left( {\dfrac{{{{\text{r}}^2}}}{4}} \right)}} \\
   \Rightarrow {{\text{F}}_1} = \dfrac{{{\text{16K}}{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}{\text{ }} \to {\text{(2)}} \\
 $
By dividing equation (2) by equation (1), we get
$
  \dfrac{{{{\text{F}}_1}}}{{\text{F}}} = \dfrac{{\dfrac{{{\text{16K}}{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}}}{{\left( {\dfrac{{{\text{K}}{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}} \right)}} \\
   \Rightarrow \dfrac{{{{\text{F}}_1}}}{{\text{F}}} = \dfrac{{{\text{16K}}{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}} \times \dfrac{{{{\text{r}}^2}}}{{{\text{K}}{{\text{q}}_1}{{\text{q}}_2}}} \\
   \Rightarrow \dfrac{{{{\text{F}}_1}}}{{\text{F}}} = 16 \\
   \Rightarrow {{\text{F}}_1} = 16{\text{F}} \\
 $

Therefore, the required force between the charges, if the distance between them is reduced to half and the charge on each particle is doubled is equal to 16F.
Hence, option D is correct.

Note: In these types of problems, it is important to note that if the two charges which are separated by a distance are like charges (i.e., either both are positive charges or both are negative charges), then there will be force of repulsion and if the charges are unlike charges (i.e., one positive charge and other negative charge) then, there will be force of attraction.