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Homework Help: Equilibrium Position of Charged Ball near Charged Plate

  1. Jul 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A small positively charged sphere of mass m is attached to a very large positively charged vertical plate by a string and hangs due to gravity. The charge density on the plate, σ, is uniform.

    a) Calculate the equilibrium position of the ball as a function of the (uniformly distributed) charge on the ball. I.e. what angle does the string make with the vertical?

    b) estimate the capacitance between the sphere and the plate.

    2. Relevant equations

    F = kqσ / r2 (not sure if this is correct)
    E = kq / r

    3. The attempt at a solution

    a) Pythagorean theorem will come in to use. I have to find where the force due to the magnetic field of the plate is equal to the force of gravity resisting it, but I can't figure out how to make the force of gravity,the force of the electric field and the tension/normal force to be in vectors that cancel out. More specifically, i can't figure out how to express the downward force of gravity to be perpendicular to the plate to cancel out the force from the plate.

    b) once i find this distance, capacitance will just be ε0*A / d with A being ∏r2 with r being the radius I would estimate the ball to be.
  2. jcsd
  3. Jul 5, 2012 #2
    Hello VictorWutang,
    For the first part the tension acts as the balancing force.Its horizontal component balances the electric force due to plate and the vertical component balances gravity.
  4. Jul 5, 2012 #3
    So is the x component of tension is sinθ * Ftension?

    and then Ftension is mass * gravity?
  5. Jul 5, 2012 #4
    Hello VictorWutang,
    The tension needs to be resolved both vertically and horizontally right?
    So Ftension is mass * gravity?[/QUOTE]
    doesn't sound correct.Would you put a picture of where you take your angle from?
    | \
    | \
    | \
    | \
    | 8
    I think the oblique slashes represent the string in your diagram the vertical slashes the plate and 8 the charge .
    Please check again.
    EDIT: Spaces were automatically truncated in my reply.So apologies for the confusing diagram .
  6. Jul 5, 2012 #5
    yes, that's what the diagram looks like. and θ is the angle i'm looking for, the angle between the wall and the rope
  7. Jul 5, 2012 #6
    Hello again,
    Then wouldn't Tsinθ=Electrical force(Calculate)
    and Tcosθ=Weight ?
    You can eliminate tension from your equations wouldn't you?

  8. Jul 5, 2012 #7
    so electrical field is E = .5kσ/r and electrical force is F = qE so

    Tsinθ = F = qE = .5qkσ/r, also Tcosθ= mg giving us T = mg/cosθ so

    mgtanθ = .5qkσ/r so

    θ = tan-1 (.5qkσ/rmg) but r in this case is the distance not the radius, and we want to put θ in terms of q being variable

    so how can i put r in terms of q?

    -----also, if you could help with the other two problems i posted i'd really appreciate it. you're a big help on this one too.
  9. Jul 5, 2012 #8
    Hello again,
    The field due to an infinitely long sheet of charge with a given charge density σ
    is given by (σ/(2ε_o))It is not a function of r.Your expression for E when revised would lend you the desired result.
  10. Jul 6, 2012 #9


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    Homework Helper

    I draw a picture with Paint and uploaded. It looks nicer than that made of characters :smile:


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