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    Physics: Find the electric field on any point on a y-axis and find the electric field as a function of the radial distance.

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

    5.) A square of paper measuring a on a side carries a total charge +q which is uniformly distributed over its surface. The square lies in the x-y plane with its center at the origin and its sides parallel to the coordinate axes. Find the electric field on any point on the y-axis with y > a. Show that this square looks like a point charge for the case y>> a.

    9.) The electric charge of the proton is not concentrated at a point but rather distributed over a volume. According to experimental investigations at the Stanford Linear Accelerator, the charge distribution of the pro- ton can be approximated by an exponential function

    e
    p= ____ e^((-r)/b)
    8(pi)b

    where r is the radial position inside the proton and b is a constant equal to 0.23 x 10^(-15)m. Find the electric field as a function of the radial distance. What is the magnitude of the electric field at r =1 x 10^(-15)m? Compare the electric field strength you find to that of a point charge of magnitude e. At what distances r do these two differ by 10% or more?

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    https://brainmass.com/physics/electric/physics-electric-field-point-axis-521237

    Solution Preview

    See the attached file.

    We begin by finding the field of a infinitely thin strip of charge (with linear charge density), of length 2L, parallel to the x axis at distance y off center.
    Each such charge equals
    (1.1)
    Each of these charges contributes an electric field of:
    (1.2)
    Where and k is Coulomb constant

    Due to symmetry, the horizontal contributions at point P from all these small charges will cancel itself and we are left with the vertical component:
    (1.3)
    Now we have to integrate all these contributions from to , but due to the parity of the integrand we need to integrate between to and multiply by 2:
    (1.4)
    To solve this integral we use the substitution:
    (1.5)
    Then:

    Putting this together:

    (1.6)
    Note that
    (1.7)
    Hence the integral becomes trivial:
    (1.8)
    And its solution is:
    (1.9)
    If we use the identity (see appendix, equation 1.38)
    (1.10)
    We get:

    (1.11)
    This is the electric field of a single thin strip of charge of length 2L at distance y off its center.

    Now we want to find the electric field along the y-axis, of a square plate of side length of 2L, centered about the origin, parallel ...

    Solution Summary

    The solution finds the electric field on any point on a y-axis.

    $2.19