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

Feb 26, 2018

$4$ times the initial

#### Explanation:

From Coloumb's law of electrostatic force we know,

$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$ (where, $k$ is a constant, ${q}_{1}$ and ${q}_{2}$ are two charges separated by distance $r$)

So,if the distance between the charges is halved, with no changes of the charges,new distance becomes $\frac{r}{2}$

So,if now force acting between the same charges is F"

Then $F ' = \frac{k {q}_{1} {q}_{2}}{\frac{r}{2}} ^ 2 = 4 \frac{k {q}_{1} {q}_{2}}{r} ^ 2 = 4 F$

So,that means, force will be $4$ times of the initial value

Feb 26, 2018

The force between the two charges is quadrupled.

#### Explanation:

The force between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Hence, if distance between charges is halved (charges remaining kept constant),

the force between the two charges is quadrupled.