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Electric field inside a cavity of a uniformly charged sphere

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Well, there are two problems, both of which would allow me to receive extra credit in my college physics class! Both are rather descriptive, so I have linked an image to it.

    http://img87.imageshack.us/img87/9570/1000497y.jpg" [Broken]

    2. Relevant equations

    For the first, I was considering the formula for electric field within a uniformly charged sphere (E= Qr/4pi[tex]\epsilon[/tex]R^3), and was thinking maybe some potential equations could come into play. (tex]\Delta[/tex]V = INT(E dl)

    For the second, I was rather clueless. The wording was ambiguous and vague, or rather I didn't under stand it :p
    I thought perhaps Force plays into this due to deflection of particles. Also, since we are given an initial velocity(? are we lol), Potential and momentum can come into play

    U = kqq/r
    Conservation of momentum.

    3. The attempt at a solution

    For the first, I couldn't quite break it down in any manner, and decided the field would be zero.

    The second, I didn't understand if they touched, or if they didn't. I thought d was the closest they could have gotten, or maybe they repel each other, and the distance with which they are closest is when their potential energy is the greatest (KE'+UE' = KE"+UE")
     
    Last edited by a moderator: May 4, 2017
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  3. Sep 22, 2009 #2

    gabbagabbahey

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    For the first problem, you can use the principle of superposition. Just calculate the individual fields (be careful, they're vectors!) of a uniformly charged sphere centered at the origin with radius [itex]R[/itex] and charge density[itex]\rho_0[/itex], and a second sphere with a center a distance [itex]L[/itex] from the origin and radius [itex]r[/itex] and charge density [itex]-\rho_0[/itex]. Then add them together (vectorially!).

    For the second problem, the impact parameter [itex]d[/itex] tells you the closest the particles would get if there was no electrostatic force deflecting them. In actuality, there are electrostatic forces (assuming you are working in the rest frame of one of the particles, the forces will be static) repelling them, and so you expect to find that the closest the particles get is actually some quantity larger than [itex]d[/itex]. You can calculate this distance using Newton's 2nd Law.
     
  4. Sep 23, 2009 #3
    Thank you for your help on the first problem, but some other student already got the credit for it.

    The 2nd problem is still up for grabs, but I don't quite understand how the velocity of the first particle will play into this. So the distance will have to be larger than the impact parameter "d", but since the moving particle is coming from infinity, how would I go about calculating the force?

    Would I have to use conservation of potential energy/momentum at all? This is still quite confusing, How would I use (F=ma) to calculate the closest distance
     
  5. Sep 23, 2009 #4

    gabbagabbahey

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    Work in the rest frame of the particle which is initially at rest in the lab frame...set up a coordinate system (I'd choose the particle at rest to be at the origin, and have the other p]article initially at [itex]x=-\infty[/itex] and [itex]y=d[/itex], with initial velocity [itex]\textbf{v}=v_{\infty}\mathbf{\hat{x}}[/itex]...so that if there were no forces acting on it, it would eventually be a distance 'd' directly above the particle at the origin)....What force acts on the moving particle? Use that to calculate the particle's trajectory, and then finally calculate the closest that trajectory will take it to the particle at the origin.
     
  6. Sep 23, 2009 #5
    The force acting on the moving particle would be (F=kQ^2/r^2), where r is the distance between the two particles. Now from there I can get the acceleration of the particle, which is the above equation for FORCE divided by mass M. So for acceleration I got [tex]\frac{kQ^2}{R^2M}[/tex]

    [tex]\Delta[/tex]Y = 1/2at^2

    V(infinity) x t =[tex]\Delta[/tex]X = infinity

    t = (infinity / V(infinity))

    plug it into [tex]\Delta[/tex]Y = 1/2at^2

    and displacement due to the force (trajectory) = (1/2)(KQ^2/r^2M)(infinity/Vinf^2)

    That is the displacement due to the trajectory @ the origin, so that plus D would yield the closest distance between the two particles...hopefully this is right. This is as close as I can get, if you can please help me out :( Very important, :/
     
  7. Sep 23, 2009 #6

    gabbagabbahey

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    Careful, force and acceleration are vectors...when the moving particle is at position (x,y), what is the force it feels? What is its acceleration?
     
  8. Sep 23, 2009 #7
    How would I go about breaking the force/acceleration into vectors just given the impact parameter d.

    As of now, I have the particle @ rest centered @ (0,0).

    The moving particle is @ (-infinity,d). The force between the two is [tex]\frac{kQ^2}{r^2}[/tex] correct? So if I wanted to find the vertical component of the acceleration (which is relevant in this context), I would need to multiply the acceleration by the sineofangleX, which is just (d/[tex]\sqrt{d^2 + infinity^2}[/tex])). infinity^2 = infinity

    So would the acceleration be [tex]\frac{kQ^2}{R^2M}[/tex] TIMES sinX which is (d/[tex]\sqrt{d^2 + infinity}[/tex])?


    Sorry If I am not getting it, lol...
     
  9. Sep 23, 2009 #8

    gabbagabbahey

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    Suppose that the moving particle is at (x,y)...what is 'r' in terms of 'x' and 'y'?...What is the direction of the force?
     
  10. Sep 23, 2009 #9
    "r" in terms of "x and y" would be[tex]\sqrt{x^2 + y^2}[/tex]

    The direction of the force is aimed upward and repelling at all times.

    Force = [tex]\frac{kQ^2}{x^2 + y^2}[/tex]

    From that, could I find the y-vector component by multiplying it by sin(x), which is (X/[tex]\sqrt{x^2 + y^2}[/tex])? After finding that y component, dividing by M would make turn it into acceleration. Correct me if I am wrong...which I probably am.

    This is somewhat difficult for me to grasp, am I headed in the right direction?
     
  11. Sep 23, 2009 #10

    gabbagabbahey

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    Right.

    Not quite. The force on the moving particle is always repulsive, but it is directed from the particle at the origin to the moving particle...so it will have two components...one vertical and one horizontal...

    That's the y-component, but what about the x-component?
     
  12. Sep 23, 2009 #11
    I just noticed, the Y component I put down was wrong. It should be Y/hyp, not X/hyp. Oops!

    Verticle component of Force = [tex]\frac{KQ^2}{X^2 + Y^2}[/tex] [tex]\frac{Y}{SQRT(X^2 + Y^2)}[/tex]

    Horizontal component of Force = [tex]\frac{KQ^2}{X^2 + Y^2}[/tex] [tex]\frac{X}{SQRT(X^2 + Y^2)}[/tex]

    Divide each component by M to find X&Y accelerations.

    Using this we can find the trajectory of the moving particle, and it's distance from Particle2 which is @ rest @ (0,0). I supposed motion equations would come into play here.

    [tex]\Delta[/tex]Y = 1/2at^2
    [tex]\Delta[/tex]X = V"t + 1/2at^2 ; where V" = Vinfinity?
    Our [tex]\Delta[/tex]X = infinity right?

    I'm still quite unsure how to relate the two accelerations/forces we have found to our trajectory. Been a while since mechanics :/
     
  13. Sep 23, 2009 #12

    gabbagabbahey

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    Right, so you can right this in vector function form as

    [tex]\textbf{a}(x,y)=\frac{kq^2}{m\left(x^2+y^2\right)^{3/2}}\left(x\textbf{i}+y\textbf{j}\right)[/tex]

    Those are the equations of motion for constant acceleration...the acceleration you just found depends on the position of the moving particle (explicitly) and since the particle's position deends on time, that means the acceleration depends on time (implicitly), and hence is not constant.

    Instead you want to use the fact that

    [tex]\textbf{a}=\frac{d^2\textbf{r}}{dt^2}=\frac{d^2 x}{dt^2}\textbf{i}+\frac{d^2y}{dt^2}\textbf{j}=\frac{kq^2}{m\left(x^2+y^2\right)^{3/2}}\left(x\textbf{i}+y\textbf{j}\right)[/tex]

    To find the particle's trajectory [itex]x(t)[/itex] and [itex]y(t)[/itex]....you'll need to solve the (coupled) differential equations...
     
    Last edited: Sep 23, 2009
  14. Sep 23, 2009 #13
    So I would have to go about integrating the functions...twice...correct?

    That would then yield the particle's trajectory, x(t) and y(t), only one of which is relevant since we are trying to find the particle's displacement above the origin...right? :(

    integrate the x component in terms of dx and the y in terms of dy?
     
  15. Sep 23, 2009 #14

    gabbagabbahey

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    You have two coupled differential equations:

    [tex]\frac{d^2x}{dt^2}=\frac{kq^2 x}{m\left(x^2+y^2\right)^{3/2}}[/tex]

    and

    [tex]\frac{d^2y}{dt^2}=\frac{kq^2 y}{m\left(x^2+y^2\right)^{3/2}}[/tex]

    You need to solve them in order to find the trajectory. You can't just integrate twice with respect to 'x' or 'y' or even 't'...you need to solve them the same way you would solve any two coupled ODE's...have you studied systems of ODE's yet?
     
  16. Sep 23, 2009 #15
    V[tex]\frac{dV}{dx}[/tex] = [tex]\frac{kQ^2}{(X^2 + Y^2){3/2}(*M)}[/tex]

    I got that to yield

    V = [tex]\sqrt{-2kQ^2 / [m SQRT(X^2 + Y^2)] +2C}[/tex]

    Where could I go from there?
    (This is x component for now, but I can just change the variables at the end to change into Y)
     
  17. Sep 23, 2009 #16

    gabbagabbahey

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    No,

    [tex]\textbf{v}\frac{dv}{dx}=\left(\frac{dx}{dt}\textbf{i}+\frac{dy}{dt}\textbf{j}\right)\frac{dx}{dt}\frac{dv}{dt}\neq\frac{kq^2}{m(x^2 + y^2)^{3/2}}[/tex]

    This is not one dimensional motion, and the ODEs are coupled...you can't solve them in this manner.

    If you had uncoupled ODEs like [itex]\frac{d^2 x}{dt^2}=f(x)[/itex] and [itex]\frac{d^2 y}{dt^2}=g(y)[/itex], then you could define [itex]v_x=\frac{dx}{dt}[/itex] and [itex]v_y=\frac{dy}{dt}[/itex] and you could solve them using this method. But your ODEs are coupled so you cannot solve them separately like this.

    No, even if you previous equation were correct (which it isn't), [itex]x[/itex] and [itex]y[/itex] are both functions of time, and so [itex]y[/itex] has some unknown implicit x-dependence and so you cannot treat it as a constant when integrating.

    What methods have you learned for dealing with coupled ODEs?
     
  18. Sep 23, 2009 #17

    gabbagabbahey

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    Actually, the ODEs are difficult to solve, and you don't really need to solve them to find the distance of closest approach....use conservation of energy instead..what is the total energy of the moving particle at [itex]t=0[/itex].... how about at some later time [itex]t[/itex], when the speed of the particle is [itex]v[/itex] and it's distance from the origin is [itex]r[/itex]?
     
  19. Sep 23, 2009 #18
    Yeah man, <-- only in calc2 here. No idea on ODE and all this higher order stuff :(

    The total energy of the particle @ t=0 is U+K = 0 + (1/2)M(Vinf^2)

    At sometime later the energy is:
    U'+K' = [tex]\frac{kQ^2}{sqrt(x^2 + y^2)}[/tex] + (1/2)M(v^2)

    So, to find when the distance is the smallest between those particles, where particle2 is @ (0,0), conservation of energy comes into play.

    Kinitial = U' + K', and where the potential energy is the greatest would be where the distance is the shortest? Am i leaning in the right direction?
     
  20. Sep 24, 2009 #19

    gabbagabbahey

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    Okay, using [itex]r=\sqrt{x^2+y^2}[/itex] you have [itex]U_0+K_0=\frac{1}{2}mv_{\infty}^2[/itex], and [itex]U(t)+K(t)=\frac{kQ^2}{r(t)}+\frac{1}{2}mv(t)^2[/itex]

    So conservation of energy tells you

    [tex]\frac{1}{2}mv_{\infty}^2=\frac{kQ^2}{r(t)}+\frac{1}{2}mv(t)^2[/tex]

    Now, you might as well solve this equation for [itex]r(t)[/itex], since this is the distance between the particles at any given time....once you've done that, what can you say about [itex]r'(t)[/itex] when the distance between particles is at a minimum?
     
  21. Sep 24, 2009 #20
    I feel so stupid right now...:/ I'm glad you're persistent and know your stuff :D

    r'(t) = 0 when the distance between the particles is a minimum.

    I found r(t)= [tex]\frac{2KQ^2}{m(Vinf^2 - V(t)^2)}[/tex]

    How exactly would I go about to find the derivative of r(t), which is r'(t)? What exactly am I deriving...LOL?
     
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