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?

 

[UNG]

Well I thought because,

Q1 = -Q2 = 100nC, then Q2 will just be -100nC
and
Q3 = -Q4 = 200nC, then Q4 will equal -200nC

Thanks for your help


I also have another question I am not too sure about.
(Click on the attachment for question and diagram)

In this next question I am guessing because the length between particle 1 and 2 is 9.00cm that particle 3 will have to be located above the two points giving an equilaterial triangle.

From this we know all the sides will be the same length of 9.00cm with an angle of 60 degrees. However, there is no solid value given for the charge on particle 1 and 2.

Since
Q1 has a charge of +Q
Q2 has a charge of +4.00Q

So does this mean that the above forumula has to be arranged somehow,
for F3,1 and F3,2 ?

 

[UNG]

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?

The force that 2 exerts on 3 would be going downwards torwards the left?

 

[learningphysics]

The force that 2 exerts on 3 would be going downwards torwards the left?

Q2 and Q3 have opposite signs right? So they will attract... so the force on 3 due to 2 is upwards and to the right.

 

[UNG]

Ok I understand that, so that makes the answer possitve.

 

[learningphysics]

In this next question I am guessing because the length between particle 1 and 2 is 9.00cm that particle 3 will have to be located above the two points giving an equilaterial triangle.

Can't be a triangle because. The 4.00q force only exerts a horizontal force on the q charge. Now placing the 3rd charge above or below the x-axis means that the net force on the q charge will have a nonzero y-component. The same thing with the net force on the 4.00q charge.

So the third charge needs to be placed on the x-axis.

 

[UNG]

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

Oh yeah that's right.
It would be by 90 degrees because its horizontal,
which means cos 90 = 0 and sin 90 = 1

Thanks

 

[UNG]

Sorry I have completed the first question,
does this look about right?

X component = -25 x 10-3 i N

y component = 46.48 x 10-3 j N

Net force = 39.18 x10-3 i j N

Are the units right? Should have a dash above the top of i and j?

 

[learningphysics]

I'm not getting those...

for the y-component I get -46.5*10^-3

For the x-component it appears you got the x-component of the force of 2 on 3 (but which should be +25*10^-3 not -25*10^-3)... but forgot to add the force of 4 on 3 which is also in the x-direction...

 

[UNG]

I see so I got the same answers, but mixed up the signs you say?

Which means that the total componets added together (Net Force) is wrong and should be a minus answer.

Therefore

Net Force = - 21.5 x10^3

?

 

[learningphysics]

No, you're missing the force of 4 on 3. please show your calculations in detail... where does the force exerted by 4 on 3 come into your calculations?

 

[UNG]

Well from the calculations it would make sense to add in a third part into the formula.

Something like this ?


<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}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{4}|}}{{{r}_{34}^{2}}\right)}{cos{\theta_{34}}}{\hat{i}}<br /> <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}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{4}|}}{{{r}_{34}^{2}}\right)} {sin{\theta_{34}}}{\hat{j}} <br /> <br />

 

[learningphysics]

Yes, that's right. Be careful about signs... show your calculations...

 

[UNG]

The answer is different since I did the Q4 and Q3.
It must be the signs.


F3x = (8.99 x10^9) x (200 x 10^-9) x (100 x 10^-9) / 0.05^2 x cos 90 i

+ (8.99 x10^9) x (200 x 10^-9) x ( -100 x 10^-9) / 0.0707^2 x cos 45 i

+ (8.99 x10^9) x (200 x 10^-9) x ( -200 x 10^-9) / 0.05^2 x cos 90 i

= 0 + -0.025 + 0 = -0.025





F3y = (8.99 x10^9) x (200 x 10^-9) x (100 x 10^-9) / 0.05^2 x sin 90 j

+ (8.99 x10^9) x (200 x 10^-9) x ( -100 x 10^-9) / 0.0707^2 x sin 45 j

+ (8.99 x10^9) x (200 x 10^-9) x ( -200 x 10^-9) / 0.05^2 x sin 90 j

= 71.92 + -25.44 x 10^-3 + - 143.84 x 10^-3 = 71.75

F3 = ( (-25 x10-3)^2 + (71.75)^2 ) Square Root
= (5148.06) Square Root
= 71.75

 

[learningphysics]

The angle between charge 3 and charge 4 is not 90... it is 0...

also, please be careful about signs... like I said before, get the magnitudes first, and then find out the sign separately by analyzing the situation... don't simply rely on the numbers to give you the right signs...

What is the direction of the charge number 2 exerts on number 3? don't use the numbers... just using the signs of the charges and the geometry you should be able to figure out the direction... is it upwards downwards to the left or to the right?

 

[UNG]

The direction of the charge that number 2 exerts on 3 will be downwards torwards the left because charge 3 is positive and charge 2 is negative. ?


The answers I get now from replacing 90 with 0 are,

F3x = 0 + - 0.025 + -143.84 x10^-3

so does this mean that -0.025 will become a +


Also for the direction of the charge that number 4 exerts on 3 will also be torwards the left because charge 3 is positive and charge 4 is negative. ?

This means that for, (when changing sin 90 to sin0)

F3y = 71.92 + -25.44 x 10^-3 + 0

-25.44 x 10^-3, Q2 on Q3 will stay negative.

Would that be it?

 

[learningphysics]

The direction of the charge that number 2 exerts on 3 will be downwards torwards the left because charge 3 is positive and charge 2 is negative. ?

charge 2 is attracting charge 3 since they are the opposite sign... if charge 3 is on the bottom left and charge 2 is on the top right... the force is upwards and to the right.

charge 1 repels charge 3 since they are the same sign... hence the force that 1 exerts on 3 is downwards...

charge 4 attracts charge 3 since they are the opposite sign... hence the force that 4 exerts on 3 is to the right.

The x-component of the force of 1 on 3 is:
0

The y-component of the force of 1 on 3 is:
-kq1q3/r^2 = 9*10^9(100*10^-9)(200*10^-9)/0.05^2 = -0.072N

So the force of 1 on 3 is: -0.072N\vec{j}

The x-component of the force of 2 on 3 is:
kq2q3/r^2*cos(45) = 9*10^9(100^-9)(200*10^-9)/0.0707^2cos(45) = 0.02546N

The y-component of the force of 2 on 3 is also: 0.02546N

So the force of 2 on 3 is: 0.02546N\vec{i} + 0.02546N\vec{j}

The x-component of the force of 4 on 3 is: kq4q3/r^2 = 9*10^9(200*10^-9)(200*10^-9)/0.05^2 = 0.144N

The y-component of 4 on 3 is 0.

So the force of 4 on 3 is: 0.144N\vec{i}

So the 3 forces are:

-0.072N\vec{j}
0.02546N\vec{i} + 0.02546N\vec{j}
0.144N\vec{i}

adding up the 3 forces gives the net force on 3:

0.16946N\vec{i} -0.04654\vec{j}

 

[UNG]

Thanks very much for your help!

I think I fully understand that question now
and have now managed to get the same answers.

Thanks

 

[learningphysics]

Thanks very much for your help!

I think I fully understand that question now
and have now managed to get the same answers.

Thanks

cool! no prob! glad you got the same answers, because I was making some mistakes, so I was a little worried that I might have messed up the answers. But I think everything's right now.