Question

If the distance between two charges is halved, what will happen to the force between the charges?

Answer 1

Here in this question, 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.

Formula used:
$F = K\dfrac{{{q_1} \times {q_2}}}{{{r^2}}}$
$F$ is the force on both the charges,
${q_1}$ charge on first particle and
${q_2}$ charge on second particle.

Complete step by step answer:
Let us assume,
Initial charge on the first particle is ${q_1}$ .
Initial charge on the second particle is ${q_2}$ .
And the initial distance between two given stationary charges is $r$ .
As we know that according to Coulomb’s inverse square law, the electric force acting between two stationary charges ${q_1}$ and ${q_2}$ when separated by a distance of r is given by:
$F = K\dfrac{{{q_1} \times {q_2}}}{{{r^2}}}$ -----(1)
Now, according to the question, the distance of the charges is reduced to half.
i.e., Final distance of separation is $\dfrac{r}{2}$ .
Now,
Let ${F_1}$ be the final electric force acting between ${q_1}$ and${q_2}$ which are separated by a distance of $\dfrac{r}{2}$ .
Now, replacing $r$ with $\dfrac{r}{2}$ and $F$ by ${F_1}$ in equation (1), we get,

$${F_1} = K\dfrac{{{q_1} \times {q_2}}}{{{{\left( {\dfrac{r}{2}} \right)}^2}}} \\\ \Rightarrow {F_1} = K\dfrac{{{q_1} \times {q_2}}}{{\dfrac{{{r^2}}}{4}}} \\\$$

$\Rightarrow {F_1} = 4K\dfrac{{{q_1} \times {q_2}}}{{{r^2}}}$ -----(2)
Which is four times the previous equation.
So, we can say that the force will be four times the initial value.

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.