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Help on Electricity and magnetism HW

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A force eld E has the spherical components
    Er = (2Dcosθ)/r3 Eθ = (Dsinθ)/r3 E[itex]\phi[/itex] = 0:

    (a) Evaluate by line integration the work it does in taking a point (parti-
    cle) from point A in the diagram to point B via the quarter circle r = a; 0 < θ < =2. (See accompanying diagram.)

    (b) Do the same for the path that includes the sector r = b < a; 0 < θ < 3[itex]\pi[/itex]/2.

    (c) You should conclude that E is derivable from a scalar potentia via
    E = -[itex]\Phi[/itex]. Find in a gauge such that [itex]\Phi[/itex](r→∞) = 0.

    (d) Show that each piece of work you evaluated in (a) and (b) equals the
    potential dierence between the endpoints of the piece.


    2. Relevant equations

    a) w=[itex]\int[/itex]E(dot product)dl

    b) i can figure out once i get a

    c)not sure how to get E from just having E=-∇0 or am i over thining this

    d) need some sort of elaboration on this part not sure what it's asking.

    3. The attempt at a solution

    MY understanding of what im doing would be a line integral so for parts a and b my guess would be to take the integral ∫aaErrdr + ∫0[itex]\pi[/itex]/2Eθrdθ + ∫E[itex]\Phi[/itex]d[itex]\Phi[/itex]

    so it would come out to
    0[itex]\pi[/itex]/2Eθrdθ which simplifies to D/r2



    also i can't find the equation used for the below question anywhere in my book just help with the equation would be great
    1. The problem statement, all variables and given/known data

    A neutral particle of mass m and a permanent dipole D (like a water
    molecule) is placed at the center of a uniformly charged ring (radius a, charge
    q > 0).

    a) What is the minimum initial velocity um that the particle must be
    given if it is to escape to infinity, tail first along the axis of the ring?

    b) What happens if the particle is disturbed with less than that minimum
    velocity?
     

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    Last edited: Nov 11, 2013
  2. jcsd
  3. Nov 12, 2013 #2

    rude man

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    It might seem like it's insufficient to just give the dipole moment, which is the product of one of the charges times distance of separation: D = qd. But, maybe if you do the math that turns out not to be the case.

    So, treat the problem as two separate charges separated by d. The direction of travel towards infinity is such that the +q charge leads the -q charge so the net force on the dipole is always towards the charged ring, and positive force needs to be done to bring the dipole to infinity.

    Note that the force on D is initially zero (at the location of the center of the charged disc.

    Good problem!

    Hint: since the E field obviously can't be uniform, would a gradient be involved in some way? :smile:
     
    Last edited: Nov 12, 2013
  4. Nov 12, 2013 #3

    rude man

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    Find out if 0 < θ < π/2 isn't meant. 0 < θ < 2 does not correspond to the figure, going from A to B along radius a.
     
  5. Nov 12, 2013 #4
    the formula i was given today by my instructor was F = (p (dot product) ∇)E

    He also said that I needed to find the potential first... now how do you take the reverse divergence as shown above for the p(dot)∇ and how would i find the potential of this by solving for E and pluging that E into V=-∫ba E dl??
     
  6. Nov 12, 2013 #5
    sorry its supposed to be 0 < θ < π/2 i must have made a mistake typing it
     
  7. Nov 12, 2013 #6

    rude man

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    The formula he gave is what I had in mind: Fx = D * grad Ex.

    I don't see why you need to get the potential. In fact, I don't see what one would do with it.

    You have an expression for Ex along the axis of the charged ring. Just go with that.

    PS oh I suppose you could calculate the E field along the axis by calculating the potential first and then taking E = - grad V. But chances are your textbook already has an expression for Ex(x).
     
  8. Nov 12, 2013 #7
    that makes since... so what would i do at this point... Fx = D * grad Ex and plug 1/2 m v2 for Ex then solve for V
     
  9. Nov 13, 2013 #8

    rude man

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    No, no. You can't equate energy with the electric field. Different units!

    You compute Fx(x) = D*grad Ex
    = Dx{grad Ex}x since we're staying on the x axis.

    Once you know Fx(x) you can integrate it and equate it to the initial kinetic energy.

    Same idea as escape velocity from the surface of the Earth. How did you compute that?
     
  10. Nov 13, 2013 #9
    sorry spent all day studying for another exam and guess i started to mix up variable ha...
    so basically i sould get Dx(∇Ex)= Fx is D a constant at this point?

    then by using the eqs
    W=∫F and W=[itex]\Delta[/itex]KE we get [itex]\Delta[/itex]KE = ∫F = initial KE

    then using what i made the mistake of doing last night finding V from the KE equation

    so i should get somthing like this (I'm assuming D is constant)

    (D∫∇E)=KE

    but the escape velocity is when the KE and the gravitational potential energy is 0 and the formula is 1/2mv2 + -Gm/r=0 so Ve= √(2Gm/r)

    or would the escape velocity in this case be KE and just electric potential energy is 0
    so
    1/2mv2+-1/(4[itex]\pi[/itex]ε0)*(Qq/r)= 0 so then
    Ve=√(-2/(4[itex]\pi[/itex]ε0)*(Qq/r)/m)
     
    Last edited: Nov 13, 2013
  11. Nov 13, 2013 #10

    rude man

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    It's ∫Fdx, not ∫F.

    Yes, D is a constant, it's given and it has an x component only.

    So your force is along the x axis and is F = D grad E. grad E is a vector pointing along the x axis and so has only an x component also.

    Your next job is to get E which is also along the x axis only. Compute it any way you want or look it up.
    Yes, if you fix your integral.
    I did not intend for you to invoke any gravitational formulas. I wanted you to think of how the escape velocity is derived without using potentials. It's derived by integrating the gravitational force from the radius of the Earth to infinity. Here your force is not gravitational but electrostatic, and you will be integrating from x = 0 to x = ∞. Don't try to mix potentials into the computations. It's beside the point.

    In both cases the integral of force over the relevant distance is equated to the initial k.e. as you say.
     
  12. Nov 13, 2013 #11
    So I will take this ( D∫∇E)=KE and solve for v by putting in 1/2mv^2 for KE how did you go from (D (dot) gradient)E to D(gradient E)
     
  13. Nov 13, 2013 #12

    rude man

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    The vector equation for the x component of force is Fx = D*grad Ex. But for you, D = D i so the vector equation reduces to D times the magnitude of grad Ex.

    D is just a constant vector. D is its magnitude. grad Ex is also a vector. This vector is a function of x. But both vectors point in the x (in the i) direction so the dot-product is simply the product of their magnitudes.
     
  14. Nov 13, 2013 #13
    ok thanks for all the help
     
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