Question

Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased?

Short answer

The separation became $\frac{1}{5}$.times of the original.

Step-by-step solution

Step 1:Given Data

Force increased by a factor of 25

Electrostatic force

The force of attraction or repulsion that exists between two-point charges separated by some distance in space is known as electrostatic force.

Comparing forces

The electrostatic force between two-point charges q and Q separated by some distance r is given as,

$F=\frac{KqQ}{r^2}..........(1.1)$

When the separation between the charges is changed to r^', the new electrostatic force is given as,

$F^{\text{'}}=\frac{KqQ}{r^{\text{'}2}}..........(1.2)$

Taking the ratio of equations (1.1) and (1.2),

$$\begin{gathered} \frac{F}{F^{\text{'}}}=\frac{{\displaystyle \frac{KqQ}{r^{\text{'}}}}}{{\displaystyle \frac{KqQ}{r^{\text{'}2}}}} \\ \frac{F}{F^{\text{'}}}={\left(\frac{r^{\text{'}}}{r}\right)}^2 \end{gathered}$$

Since $F^{\text{'}}=25F$. Therefore,

$$\begin{gathered} \frac{F}{25F}={\left(\frac{r^{\text{'}}}{r}\right)}^2 \\ {\left(\frac{r^{\text{'}}}{r}\right)}^2=\frac{1}{25} \\ \frac{r^{\text{'}}}{r}=\frac{1}{5} \\ {r}^{\text{'}}=\frac{r}{5} \end{gathered}$$

Hence, the separation between the charges became $\frac{1}{5}$of the original.