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    Working with an electroscope

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    Two conducting wires each of length L are connected to a knob, whereby a total electric charge Q is deposited onto two small spherical metallic spheres, each of mass m and diameter d, at the end of the wires. We observe each wire of this electroscope to lie at an angle q from the vertical.

    Find an expression for the charge Q in terms of the given quantities L,m,d and q as well as physical constants. You have now calibrated the electroscope, and may use it for quantitative experiments.


    Compute the electric field, in both components and as magnitude and direction, at the midpoint between the two hanging masses

    © BrainMass Inc. brainmass.com December 24, 2021, 4:52 pm ad1c9bdddf

    SOLUTION This solution is FREE courtesy of BrainMass!

    Please view the figure also

    When the electroscope is charged, an equal amount of charge is transferred to each of the balls, which repel each other causing the leaves to make an angle with the vertical.
    Thus the horizontal forces keeping the balls apart and hence the leaves are the Coulomb forces due to each charged ball. Also, we have to remember that there will consequently be some tension in the leaves of the electroscope which are held at an angle to the vertical. In addition, we have to take into account the weight of the balls as they have a finite mass. Thus there are a number of forces acting on the system which need to be resolved in this problem.

    The electrostatic force between the two balls, each with a charge q = Q/2, is given by,

    F = K charge^2/R^2 [where k = [1/4 pi e0 ]= 9*10^9]

    But we have charge = Q/2 and from the figure, we get R = 2r substituting back,

    F = K (Q/2)^2/(2r)^2 = K Q^2/16r^2 ..............(1)

    At equlibrium, Sum [Fx] = 0 (no net force in the x direction)

    Thus T Sin(q) = F (force in x direction)

    where T sin(q) is the x component of the tension.

    thus we have, T Sin(q) = F (force in x direction) .....(2)

    Similarly, T Cos(qa) = mg .....(3)

    From 2 and 3, we get F = mg tan(q) ......(4)

    from 4 and 1, we will write,

    mg tan(a) = K Q^2/16r^2 ---------(basic equation)...(5)

    from this Q = sqrt [mg tan(q)*16 r^2/K]

    from the figure, sin(q) = r/L ====> r = L sin(q)

    Q = sqrt [mg tan(q)* 16 (L Sin(q))^2/K]

    Q = 4L sin(q)* sqrt[mg tan(q)/K]------(6)

    now, we have the charge on each ball (same charge)

    Since the charges are the same, the magnitude of the electric field produced by the balls are equal but in opposite directions. At the mid point, the field vanishes.

    Electric field intensity at a distance s from a ball is

    E = K Q/s^2

    best wishes...

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:52 pm ad1c9bdddf>