Question

What happened to electrostatic force if both charges and distance is doubled

Answer 1

Let initial charges be q and Q. And the separation distance be r

Let initial force be F .

$F = \dfrac{1}{4\pi \epsilon} \dfrac{q Q}{ {r}^{2} }$

As per the question , both the charges as well as the separation distance has been doubled :

So new Charges are 2q and 2Q ,

New separation distance be 2r

Let new Force be F2

$F2 = \dfrac{1}{4\pi \epsilon} \dfrac{(2q)(2 Q)}{ {(2r)}^{2} }$

$= > F2 = \dfrac{1}{4\pi \epsilon} \dfrac{4qQ}{ 4{r}^{2} }$

$= > F2 = \dfrac{1}{ 4\pi \epsilon} \dfrac{ \cancel4qQ}{ \cancel4{r}^{2} }$

$= > F2 = \dfrac{1}{4\pi \epsilon} \dfrac{q Q}{ {r}^{2} }$

$= > F2 = F$

So the electrostatic force remains same

Answer 2

Explanation:

let ,initial charge be q

distance =r

electrostatic force=q.q/4π€•r²=q²/4π€•r²

if charge is doubled then, charge =2q

if distance is doubled then,distance=2r

electrostatic force=2q .2q/4π€•(2r) ²

=4q²/16π€•r²

=q²/4π€•r²

so, electrostatic force will be remain same.

hope it will help you.