# Electricity Physics Problem involving Coloumb's Law

1. Feb 5, 2010

### jetsfan101202

1. The problem statement, all variables and given/known data

In Fig. 21-29, particles 1 and 2 of charge q1 = q2 = +4.80*10^-19 C are on a y axis at distance d = 34.0 cm from the origin. Particle 3 of charge q3 = +8.00*10^-19 C is moved gradually along the x axis from x = 0 to x = +5.0 m. At what values of x will the magnitude of the electrostatic force on the third particle from the other two particles be (a) minimum and (b) maximum? What are the (c) minimum and (d) maximum magnitudes?

2. Relevant equations

F=Kq1q2/r^2

3. The attempt at a solution

I'm absolutely lost on this problem. I know I will have to take the derivative at some point but before that I'm lost. Any help is greatly appreciated. Thanks

2. Feb 5, 2010

### tiny-tim

Welcome to PF!

Hi jetsfan101202! Welcome to PF!

(try using the X2 and X2 tags just above the Reply box )

Start by writing the equation for the total force on particle 3 at a general position x …

what do you get?

3. Feb 5, 2010

### thebigstar25

As a start lets simply put what information we have in a figue .. here is one :

http://img20.imageshack.us/img20/9920/63571485.jpg [Broken]

the question is asking to find the positon x (between point 0 to 5 m) where the electrostatic force is min. and max >> I suggest here to find the electric field at point 0 and point 5 then apply F= Q3*E (E here is the net field produced by Q1 and Q2 ,note that you need to do vector addition).. what do you notice? and to convince yourself take some points(i.e x= 1 , x=2, x=3) what do you notice for the magnitude of the electrostatic force as you go from point x=0 (through the points i suggested) to point x=5? ..

try it this way , and if you still have problems ask again..

Last edited by a moderator: May 4, 2017
4. Feb 5, 2010

### jetsfan101202

F=Q3*E so

(8*10^-19)*(kQ1/sqrt((.342+x2)2)+kQ2/sqrt((.342+x2)2)

then just find the value of x where it is the greatest?

5. Feb 5, 2010

### tiny-tim

HI jetsfan101202!

(have a square-root: √ )

No, the electric forcs are vectors, so you can't just add the magnitudes.

6. Feb 5, 2010

### thebigstar25

keep in mind that when you deal with E the electric field you are dealing with vector quantity .. so you need first to identify the direction of the elecrtic field of Q1 and Q2 , and since both of them have positive charges then the electric field is outward (as I clarify in the figure) ..

note another thing, you have both Q1 and Q2 have the same amount of charge , this tells you that the magnitude of their electric field produced would be identical but the direction would be difference ( you can easily see that by a quick look at the figure) ..

Is it clear enough now? , well I hope it is ..

7. Feb 5, 2010

### jetsfan101202

(8*10^-19)*(-sin(theta)(kQ1/sqrt((.342+x2)2)+sin(theta)(kQ2/sqrt((.342+x2)2)))

is that correct since the E of Q1 would be down and E of Q2 would be up? How would I find theta though?

8. Feb 5, 2010

### jetsfan101202

sorry I forgot the x component of that

(8*10^-19)*(cos(theta)(kQ1/sqrt((.342+x2)2)+cos(theta)(kQ2/sqrt((.342+x2)2)))

9. Feb 5, 2010

### tiny-tim

You don't need θ, you only need sinθ …

so use sin = opp/hyp.

10. Feb 5, 2010

### jetsfan101202

so for the vertical component...

(8*10^-19)*-(d2/sqrt(x2+d2))(kQ1/sqrt((.342+x2)2)+(d2/sqrt(x2+d2))(kQ2/sqrt((.342+x2)2)))

and the horizontal...

(8*10^-19)*(x2/sqrt(x2+d2))(kQ1/sqrt((.342+x2)2)+(x2/sqrt(x2+d2))(kQ2/sqrt((.342+x2)2)))

but how do I solve for maximizing x?

simplifying the vertical compnent ends up equaling zero? is that correct

the horizontal comes out to be

Q3((2x2Q)/(x2+d2))

do I just need to maximize the function above?

11. Feb 5, 2010

### thebigstar25

:( I believe that you dont need to confuse yourself this way , I have made the figure in a previous reply so you can solve your problem in an easy striagth way :( ..

hint : have another look at the figure and notice that you have symmetry argument you can take advantage of ..

12. Feb 5, 2010

### jetsfan101202

well Q1 =Q2

so looking at your fist post you said F=Q3*E(E being the added electric field of Q1 and Q2) so

I think the issue is how do I find "E here is the net field produced by Q1 and Q2 ,note that you need to do vector addition)"

im sorry i'm getting confused. I realize that Q1 and Q2 are equally far apart from Q3 so they're veritcal electric fields will cancel and the horizontal will just be double the amount on one, but I dont know how to maximize that.

13. Feb 5, 2010

### thebigstar25

okay I can see that you are confused alittle so I will try to simplfy it as possible as I can ..

you realize that Q1 and Q2 have the same charge and and equally apart from Q3 ..

so, in order to solve this problem the easy way is to calculate the electric field produced from each charge Q1 and Q2 ..

I will help you to start with it .. lets start to notice what will be the situation if we we want to figure out what is the electric field at point x = 0, as the following figure :

http://img522.imageshack.us/img522/9091/34630616.jpg [Broken]

what do you notice? .. you will notice that the elecrtic field produced from Q1 and Q2 at the point x=0 is purely in the y-direction and each of equal magnitude but opposite direction , so what happens? as a result of that they cancel each other then the electric field at x = 0 is simply 0 , what that tells you? since F =Q3*E you can easily see that at that point F= Q3*0 = 0 ..

now, what happens when you want to find what is the electric field at a point not x=0 , say x=1 or x=2 .. you will notice that the electric field at such points is not purely at the y-direction anymore, but it has now x- and y- directions.. and as you stated always the y-directions will cancel and the horizontal will double..

another thing you should note , as you move from x=0 to x=5 the magnitude of the electrostatic force will increase (you can easily verify it by computing the force)..

I hope what im saying is clear now .. if you still dont get it , ask again..

Last edited by a moderator: May 4, 2017
14. Feb 5, 2010

### jetsfan101202

thanks again for the help. After reading over you last post it's very clear that the answer should be x=5 because it will keep increasing and 5 is the limit. The issue is that when I type 5 into the answer for "b) position of maximum force" on the webassign it is incorrect. Maybe the programming on the question is wrong?

15. Feb 5, 2010

### thebigstar25

can u find the force at points x=1 , x=2 , x=3 , x=4 and x=5 .. See if the force is really increasing when u r moving to x =5 .. Please check is .. The only thing you have to find is cos theta , and your costheta is 0.34/sqrt( (0.34*0.34)+ x^2) ..

16. Feb 5, 2010

### thebigstar25

please dont take my suggestions as final answers , try to verify them to yourself im talking from previous knowledge and i may do mistakes ,but i hope im not .. So please try it first yourself and see what results you are getting ..

17. Feb 5, 2010

### jetsfan101202

yes i did

2(.34/sqrt(.342+x2)*(kQ/(x2+.342))

and it increased from x=1 to x=2 ... x=5

thats the right equation though correct?

18. Feb 5, 2010

### tiny-tim

No, you've used cosθ = .34/ √(x2 + .342).

You should have used sinθ.

19. Feb 5, 2010

### thebigstar25

Yes, your vertical component is E costheta and the horizontal is E sintheta