Question

If thedistance between two equal point charges is doubled and thier individual charges are also doubled what happens to the force between them​

Answer 1

Given:

Distance as well as initial charges has been doubled.

To find:

Changes in the Electrostatic force between them

Calculation:

Let initial force be F ; charges be q1 and q2 and distance of separation be d .

$F = \dfrac{1}{4\pi \epsilon} \dfrac{(q1)(q2)}{ {d}^{2} }$

Now , new charges are 2(q1) and 2(q2) and new distance is 2d . Let new Force be F2 :

$F2 = \dfrac{1}{4\pi \epsilon} \dfrac{2(q1) \times 2(q2)}{ {(2d)}^{2} }$

$= > F2 = \dfrac{1}{4\pi \epsilon} \dfrac{ \cancel 4(q1)(q2)}{ { \cancel 4(d)}^{2} }$

$= > F2 = \dfrac{1}{4\pi \epsilon} \dfrac{(q1)(q2)}{ {d}^{2} }$

$= > F2 = F$

So the Electrostatic force remains same even after those specific changes in charge and distance of separation.