Question

Can someone please check my work on this equation so far? Here is the question...

Two point charges each carrying a charge of + 4.5 E – 6 C are located 4.5 meters away from each other. How strong is the electrostatic force between the two points and is this force a repulsive force or an attractive force (k = 9.0 E9 Nm2/C2)?

So the values I have are as follows...
F = ?
k = 9.0 E9 Nm^2/C^2
q1 = 4.5 E - 6 C
q2 = 4.5 E - 6 C
r = 4.5 m

Am I on the right track? Thanks!

Answer 1

@SarahCatherinez yes you have got the values of each of the parameter correct. Seem to be on the right track. Go on and plug in the values in the formula for force.

Answer 2

Great! I'll go ahead and plug in the values, if you wouldn't mind checking my work from there.

Answer 3

yes, no problem

Answer 4

F = (9.0 E9) * (4.5 E -6) * (4.5 E -6)/(4.5^2)
F = (9.0 E9) * (4.5 E -6) * (4.5 E -6)/(20.25)
F = (9.0 E9) * (4.5 E -6) * (2.2 E -7)
F = .00891 N

Answer 5

There seems to be a mistake in 3rd step.Could you check?

Answer 6

Hm, that's odd. I just entered it into my calculator and got the same answer.

Answer 7

yes, sorry. Your answer is close to the correct. I would show you another way
you have this
\[F=\frac{9*10^9*(4.5 *10^{-6})^2}{4.5^2}\]
\[F=\frac{9*10^9*(4.5)^2 *(10^{-6})^2}{4.5^2}\]
\[F=\frac{9*10^9*\cancel{(4.5)^2} *(10^{-6})^2}{\cancel{4.5^2}}\]
\[F=9*10^9*10^{-12} N\]

Do you follow this method?

Answer 8

Oh yes, that makes sense.

Answer 9

I think you can find the answer now, can't you?

Answer 10

I believe so. Thank you so much for your help! :)

Answer 11

No problem