Question

Can someone please walk me through solving this problem?

Answer 1

Two point charges exert a coulomb force on one another of F a distance sqrt(2)D apart. At what distance will the force double?

Answer 2

F = q1q2/d^2

Answer 3

I guess if Force doubles, distance decreases by 1/2?

Answer 4

1/4

Answer 5

That's a good guess, but you have missed one little thing. It is proportional to the SQUARE of the distance. This is why you changed to 1/4. You squared 1/2.

F = q1*q1/[(sqrt(2)*D)^2] Is that where we are starting?

Find
2*F = q1*q1/[(sqrt(2)*D)^2] Is that where we are starting?
Solve for F and put that "2" properly under the radical in the denominator on the other side.

Answer 6

\[\[F = \frac{ q_1 q_2 }{ 4D^2 }\]\]

Answer 7

Hint: \(4D^{2} = (2D)^{2}\)

Answer 8

:D

Answer 9

So at distance 2D

Answer 10

wait

Answer 11

I don't understand my professor's process. :/

Answer 12

Prof. Did exactly what I wanted you to do.

\(F = \dfrac{Stuff}{\left(\sqrt{2}D\right)^{2}}\)

This is where we start. Now, we wonder about the distance that would create 2F

\(2F = 2\dfrac{Stuff}{\left(\sqrt{2}D\right)} = \dfrac{Stuff}{\dfrac{1}{2}\left(\sqrt{2}D\right)^{2}} = \dfrac{Stuff}{\left(\dfrac{\sqrt{2}D}{\sqrt{2}}\right)^{2}} = \dfrac{Stuff}{D^{2}}\)