Question

Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force between them is 4N. What will the magnitude of this force become if the distance of separation were to double between them?

Answer 1

Answer:

From the formula of force:

$F = \frac{kAB}{ {r}^{2} }$

since AB and k are constants:

$F \: \alpha \: \frac{1}{ {r}^{2} } \\ \\ F = \frac{x}{ {r}^{2} }$

x is a constant of proportionality

• when force is 4N, separation distance is 1

$4 = \frac{x}{1} \\ x = 4$

therefore, equation becomes

$F = \frac{4}{ {r}^{2} }$

when r is doubled, r becomes 2. find F:

$F = \frac{4}{ {2}^{2} } \\ \\ F = \frac{4}{4} \\ \\ { \underline{force \: is \: 1N}}$