Physics

Negative charge is distributed uniformly around three-quarters of a circle of radius, a, with the center of curvature at the origin.
a) Use integration to find an algebraic expression for the x and y components of the net electric field at the origin in terms of the linear charge density lambda and the radius a.
b) Find an algebraic expression for the magnitude and direction of the electric field at the origin.
c) If the distribution has a total charge Q= -25uC and if a=5.0cm, find the numerical value for the electric field's magnitude at the origin.

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  1. You can integrate over the 3/4 circle, but think on superposition.

    If you had a full circle of linear density of -25uC/(3/4*PI*10), the E would be zero.

    So if you just took the negative of E (in both directions) for a quarter circle, wouldn't that be the same as E from a 3/4 circle, since a full circle would be zero E?
    For the quarter circle....
    Q is distributed uniformly around an arc of length pi*a/2, so an element of arc ds has charge dQ = Q/(pi*a/2)ds = 2Q/(pi*a)ds but ds = a*d(theta) so dQ = (2Q/pi)d(theta). Away you go, just as you stated.

    (I got Ex = Ey = 2*k*Q/(pi*a^2)
    and you take the negative of that.

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    bobpursley

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