Coulomb's Law and charged particles

[UNG]

1. Homework Statement
Hi, I would appreciate it if someone could help me with this question, I am a n00b here
Please and thank you
Okay here's the question.

The particles have charges Q1 = -Q2 = 100nC and Q3 = -Q4 =200nC
and distance a = 5.0cm.

What are the x and y components of the net electostatic force on particle 3 ?

So basically a diagram is given with the question showing the 4 particles equally separated from each other at a distance of 5.0cm, making a square. And I have to find the force on particle 3.

2. Homework Equations

This is the equation I used.

F1,3 = (8.854x10^-12) x (Q1 x Q3)/0.0707xR^2 (0.0707 comes from Phyagrous therom)
H= (square root) 0.05^2 + 0.05^2
F1,4 = (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2

Fnet = F1,3 + F1,4

= F1,3 + (8.854x10^-12) x (Q1 x Q3)/R^2 Cos 45 (45 degrees middle of square)
+ (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2 Sin 45

3. The Attempt at a Solution

I have a feeling this is seriously wrong and have made it more complicated than it is.

I placed the values into the equation shown and came out with

= -79.46 x10^3 + 79.46 x10^3

Heres a site I found afterwards...
http://physics.bu.edu/~duffy/PY106/Charge.html

I think it might go a little something like that instead^^

If you could help that would be great thanks.

 

[learningphysics]

I'm confused by the "pythagorean theorem" part...

What is the distance between charge 1 and charge 3?

 

[UNG]

Doesnt say.
But I worked it out by spliting the square in two (a triangle)
and used pythagoras theorem to find the longest side (diagonal between Q 1 and 3)

 

[UNG]

As in h^2 = a^2 + b^2.

So the distance between 1 and 3 should be 0.0707

 

[learningphysics]

Doesnt say.
But I worked it out by spliting the square in two (a triangle)
and used pythagoras theorem to find the longest side (diagonal between Q 1 and 3)

Why do you have 0.0707*R^2, instead of just (0.0707)^2?

The distance between 1 and 3 is... sqrt(0.05^2 + 0.05^2) = 0.0707

so F1,3 =

F1,3 = (8.854x10^-12) x (Q1 x Q3)/(0.0707)^2

And also the distance between 1 and 4 is 0.05m right? so how is the 0.0707 coming into that part?

 

[learningphysics]

As in h^2 = a^2 + b^2.

So the distance between 1 and 3 should be 0.0707

yes. so why the 0.0707*R^2?

 

[UNG]

Oh yes sorry, realized that a while ago but needed to edit it.
Also I only put the 0.0707 bit in F1,4 not F1,3

 

[learningphysics]

Oh yes sorry, realized that a while ago but needed to edit it.
Also I only put the 0.0707 bit in F1,4 not F1,3

let's start fresh... what is the magnitude of the force that particle 1 exerts on particle 3?

 

[UNG]

I worked it out as -71.92x10-3

 

[learningphysics]

wait... you're making a mistake with the 8.854x10^-12.

force = kQ1Q2/r^2,where k = 9*10^9.

instead of k you can also use: 1/(4*pi*(8.854*10^-12))

 

[UNG]

I see,
it really should be 8.99x10^9 when using the 1/ 4x pi x8.854*10^-12

 

[learningphysics]

I see,
it really should be 8.99x10-12 when using the 1/ 4x pi x8.854*10^-12

not -12... k = 8.99*10^9.

k=1/(4pi*8.854*10^-12) (the 8.854*10^-12 is in the denominator).

 

[UNG]

My mistake, another typing error

 

[learningphysics]

My mistake, another typing error

no prob. I'm getting the force between 1 and 3 as having a magnitude of 0.036N.

 

[UNG]

I did,

F1,3 = (8.99^9) x (100x10^-9)(200x10^-9)/ 0.05^2

= -71.92x10-3 N

Wrong Q 1 and 3 values I put into the equation perhaps?

 

[learningphysics]

I did,

F1,3 = (8.99^9) x (100x10^-9)(200x10^-9)/ 0.05^2

= -71.92x10-3 N

Wrong Q 1 and 3 values I put into the equation perhaps?

you should have 0.0707^2 not 0.05^2.

 

[UNG]

Oh no.
Its Q1 to Q3 are the charges on the same line.

Q1 to Q4 are diagonal, being 0.0707^2

 

[learningphysics]

Oh no.
Its Q1 to Q3 are the charges on the same line.

Q1 to Q4 are diagonal, being 0.0707^2

Ah... ok. so the magnitude of the force of Q1 on Q3 is 71.92x10-3N. Is this in the y-direction or x-direction?

Use \vec{i} and \vec{j} vectors... to write this force in vector form... careful about directions and signs.

 

[UNG]

Well Q1 and Q3 are together on the y axis
and the question asks for the force on Q3

 

[learningphysics]

Well Q1 and Q3 are together on the y axis
and the question asks for the force on Q3

Is the force upward or downward?

what are the positions of the 4 charges. Is Q1 the upper left, Q3 bottom left...

 

[UNG]

Yes,
Q1 is the upper left
and Q3 is the bottom left.

Q2 top right, Q4 bottom right

 

[learningphysics]

Yes,
Q1 is the upper left
and Q3 is the bottom left.

Q2 top right, Q4 bottom right

So the force of Q1 on Q3 is -71.92*10^{-3}N\vec{j}

What is the force that Q2 exerts on Q3? first find the net force... then get the components.

 

[UNG]

Would it be


F2,3 = (8.99x10^9) x (-100x10^-9)(200x10^-9) /0,0707^2

=-35.97 N ?

 

[learningphysics]

Would it be


F2,3 = (8.99x10^9) x (-100x10^-9)(200x10^-9) /0,0707^2

=-35.97 N ?

careful about your powers of 10... it should be 35.97*10^(-3).

So what is this force written in vector form... ie in the form F_x\vec{i} + F_y\vec{j}?

 

[UNG]

Not sure but I calculated it by using the fomula in the other thread.

F3 = F3,1 + F3,2

= (8.99x10^9) (200x10^9) (100x10^9)/0.05^2 cos 45 i = 5.086 x10^34

+ (8.99x10^9) (200x10^9) (-100x10^9)/0.0707^2 cos 45 j = -2.544 x10^34

= 5.086 x10^34 + - 2.544 x10^34

= 2.542 x10^34

 

[PFStudent]

Hey,

Do not create the same post twice in this forum.

You posted this same question yesterday.

https://www.physicsforums.com/showthread.php?t=187894

So do not repost it here again, just wait for someone else to respond or you reply with some more work, but do not just create a duplicate post.

Thanks,

-PFStudent

 

[UNG]

Yeah sorry about that,
I posted two but couldn't delete the other.

Thanks for the correct forumla, is that all I need to do for this question?

 

[PFStudent]

Hey,

No problem for the double post (actually I had a similar problem with my very first post here on PF).

Ok, so you already know

<br /> \sum{\vec{F}_{3_{}}} = {\vec{F}_{31_{}}} + {\vec{F}_{32_{}}}<br />

So, if you have already done the x-components, given below.

<br /> \sum{\vec{F}_{3_{x}}} = {\vec{F}_{31_{x}}} + {\vec{F}_{32_{x}}}<br />

<br /> \sum{\vec{F}_{3_{x}}} = {\left(\frac{{k_{e}}{|q_{3}|}{|q_{1}|}}{{{r}_{31}}^{2}}\right)}{cos{\theta_{31}}}{\hat{i}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{2}|}}{{{r}_{32}}^{2}}\right)}{cos{\theta_{32}}}{\hat{i}}<br />

Then, just work the y-components as given below,

<br /> \sum{\vec{F}_{3_{y}}} = {\vec{F}_{31_{y}}} + {\vec{F}_{32_{y}}}<br />

<br /> \sum{\vec{F}_{3_{y}}} = {\left(\frac{{k_{e}}{|q_{3}|}{|q_{1}|}}{{{r}_{31}}^{2}}\right)}{sin{\theta_{31}}}{\hat{j}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{2}|}}{{{r}_{32}}^{2}}\right)}{sin{\theta_{32}}}{\hat{j}}<br />

Finally you recognize that,

<br /> \sum{\vec{F}_{3_{}}} = {\vec{F}_{31_{}}} + {\vec{F}_{32_{}}}<br />

So then the magnitude of the net force on particle three, {\left|\sum{\vec{F}_{3_{}}}\right|} is given as follows,

<br /> {\left|\sum{\vec{F}_{3_{}}}\right|} = {\sqrt{{{\left(\sum{{F}_{3_{x}}}\right)}^{2}}+{{\left(\sum{{F}_{3_{y}}}\right)}^{2}}}}<br />

Thanks,

-PFStudent

 

[UNG]

Ah thanks very much :)
I remember doing something similar to this before with R-L circuits e.t.c

 

[learningphysics]

Not sure but I calculated it by using the fomula in the other thread.

F3 = F3,1 + F3,2

= (8.99x10^9) (200x10^9) (100x10^9)/0.05^2 cos 45 i = 5.086 x10^34

you shouldn't be multiplying by cos45 here... the force between 1 and 3 is vertical... so there is no horizontal component.

+ (8.99x10^9) (200x10^9) (-100x10^9)/0.0707^2 cos 45 j = -2.544 x10^34

= 5.086 x10^34 + - 2.544 x10^34

= 2.542 x10^34

be careful about signs... also your powers of 10... nano is 10^(-9) not 10^9.

deal with the magnitudes first... then look at the charges to find the direction... don't assume the charge signs will give you the right direction...

what is the direction of the force that 2 exerts on 3? just generally.. don't use numbers... is it to the left or to the right? is it upwards or downwards?