Question

When the distance between the charged particles is halved, the force between them becomes

• A. One-fourth
• B. Half
• C. Double
• D. Four times

Answer 1

The correct option is D

Four times

Step 1- Given data and assumptions:

First charge particle is assumed as $Q_1$

The second charge particle is assumed as $Q_2$

Distance between the charge particle is assumed as $r$

Step 2- Formula used:

According to coulomb's law, the electrostatic force between the two charged particles is directly proportional to the product of the magnitude of the charge and inversely proportional to the square of the distance between them.

The electrostatic force $F$ between the two charged particles having charge $Q_1$ and $Q_2$ and the distance between them $r$ is given by $F=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{r^2}$,where ${\epsilon }_0$is the permittivity of free space.

Step 3- Calculations:

When the distance between the charge particles is halved i,e, $r^{\text{'}}=\frac{r}{2}$, and the charge particle magnitude remains the same the new electrostatic force is given by $$\begin{gathered} F^{\text{'}}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{{r^{\text{'}}}^2}But{r}^{\text{'}}=\frac{r}{2} \\ {F}^{\text{'}}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{{r^{\text{'}}}^2}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{{\left({\displaystyle \frac{r}{2}}\right)}^2}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_1{Q}_2}{r^2}x4=4F \end{gathered}$$

So, the new electrostatic force is increased to $4$ times its initial value.

Hence, option (D) is correct.