Charge distribution, 3 charges

[Dodsy]

Homework Statement

A triangle is given with the points:

Q1 = +2.0 x 10-5 C

2.0 m from

Q3 = -3.0 x 10-5 C

AND

Q1 = +2.0 x 10-5 C

2.0 m from

Q2 = -3.0 x 10-5 CThe triangle is a right triangle, with Q1 at the 90 degree angle.FIND THE NET FORCE OF CHARGE 1

Homework Equations

[/B]
FE21 = FE31 = kq1q2 /r2
Fnet = SqrtFE21^2 + FE31^2

The Attempt at a Solution

[/B]
Because the charges are an equal distance away:

FE21 = FE31 = kq1q2 /r2FE21 = FE31 = (9.0 x 109 Nm2/C2)(2.0 x 10-5 C)(3.0 x 10-5 C) / 2.0 mFE21 = FE31 = 90’000 NFnet^2 = 90'000^2 + 90'000^2

Fnet = 127279.22 NNow I just need to calculate the angle. I just want to make sure I'm on the right track.

 

[Dodsy]

Any help would be appreciated...

 

[haruspex]

FE21 = FE31 = (9.0 x 109 Nm2/C2)(2.0 x 10-5 C)(3.0 x 10-5 C) / 2.0 m
FE21 = FE31 = 90’000 N

Not following your arithmetic here.
Did you forget the 2 in /r²?
I assume 109 means 10⁹, etc.
Is your 90'000N result 9*10⁴N?
(If you can't be bothered to use superscripts for the powers of 10, use the E notation, like 9E9.)

 

[Dodsy]

I must have! And sorry I didn't notice as I copied and pasted from the microsoft word document I am working in! I appreciate the response. Other than that is everything looking okay?

 

[haruspex]

Other than that is everything looking okay?

No, I have no idea how you get that number. I get a number less than 10.

 

[Dodsy]

Yes the answer will obviously be different now. Other than that is everything okay with my formula? I'd rather not fix my work just to have someone else copy it. If squaring my radius is the only problem that's easily fixable.

 

[haruspex]

Yes the answer will obviously be different now. Other than that is everything okay with my formula? I'd rather not fix my work just to have someone else copy it. If squaring my radius is the only problem that's easily fixable.

I'm saying that will only make a factor of 2 difference. 90000 seems to be wrong by several orders of magnitude.

 

[Dodsy]

Please tell me where I'm going wrong with my equation then? If I square my radius I get 1.35. What am I missing here? that is the purpose of this thread. As I've already stated 90000 obviously is NOT the answer. I need constructive criticism and a level of intelligence to be met with my answer, let's not go around in circles here.

 

[gneill]

It would seem that you have arithmetic errors (calculator operation problem?) to resolve.

Start by pulling apart the scientific notation in your force equation to see if the order of magnitude is in the right ballpark. You should be able to do the estimation in your head:

$10^9 \times 10^{-5} \times 10^{-5} = ~?$

and then the rest:

$\frac{9 \times 3 \times 2}{2^2} = ~?$

Does your result of 90000 agree with the above?

 

[haruspex]

Please tell me where I'm going wrong with my equation then?

You don't seem to have understood my posts #3 and #5. I quoted two lines from your OP, and said that I could not follow the arithmetic. I.e. there's a mistake in going from the first line to the second. But I have no idea what that mistake is because I cannot see the intermediate steps.
In post #5 I restated that despite picking up the distance-squared error I still had no idea how you got 90000. Yet again you came back with "is anything else wrong?" (i.e. other than the distance-squared error).

 

[Dodsy]

Once again, I know 90000 is not the answer. The answer was in fact 2.7 when I computed it. A calulator issue is NOT what I'm looking for. I'm looking for arithmatic issues. If you can't follow a simple conversation please don't bother answering my question.

FE21 = FE31 = kq1q2 /r2 FE21 = FE31 = (9.0 x 10⁹ Nm2/C2)(2.0 x 10^-5C)(3.0 x 10^-5 C) / 2.0 m² FE21 = FE31 = 1.35

Please next time, use your brain. I've taken into account the calulation issue and the radius not being squared. I want to make sure that there is nothing else wrong with this. WE HONESTLY DO NOT NEED TO KEEP GOING IN CIRCLES. I AM NOT RETARDED.If 90000 is not the answer its not the answer, move on. The answer to the simple question as to whether my formula is correct would suffice.Honestly, not impressed with this community.

 

[haruspex]

The answer was in fact 2.7 when I computed it

Then why, oh, why did you not say so before? As far as I could tell, you were now getting 90,000/2 = 45,000. You originally had 90,000, and the only error you had acknowledged was the factor of 2. So from where I sat, yes, you still had something wrong. I can only work with the information you provide.