Question

Check w this question:
If you increase the distance between two identical point charges so that the new distance is five times the original distance, what happens to the force between them?

It is multiplied by 5.

It is multiplied by 25.

It is divided by 25.

It is divided by 5.

i think divided by 5 is the answer but im not sure

Answer 1

If r=distance between the two charges, q1= value of the first charge and q2 = value of the second charge, the electrostatic Coulomb's law for the (electrostatic) Force is : \[E=k _{e}(q _{1}q _{2}/r ^{2})\]
(ke is a constant)
You have just to understand what happened on E when \[r \rightarrow 5r\]

Answer 2

@Pierrot i tried plugging in numbers but i cant get it to work

Answer 3

Ok, you have the initial force \[E _{i}=k _{e}(q_{1}q_{2}/r _{initial} ^{2})\]
If \[r=5r _{initial}\]You have now
\[E=k _{e}(q_{1}q_{2}/(5r_{initial} )^{2} \rightarrow E=k _{e}(q_{1}q_{2}/?(r_{initial} )^{2})=E _{i}/? \]

Answer 4

@Pierrot so it's divided by 25

Answer 5

Yes, that's right.

Answer 6

thank you so much for writing that out because it actually made me understand it! :)

Answer 7

You're welcome