Two curved plastic rods form a circle of radius = 8.50 cm in an xy plane. The to

Question

Two curved plastic rods form a circle of radius = 8.50 cm in an xy plane. The top half of the circle has a charge of +q, while the bottom half as a value of -q, both of which are distributed uniformly. The x axis passes through both of the the connecting pints. given q=15.0pC,, determine the magnitude and direction (relative to the positive x axis) of the electric field E produced at a height of 15.0 cm above the origin

so far I know that direction will be in postive vertical direction, but my problem is trying to figure out how to set up the integral with proper limits

Explanation / Answer

The question is about electric field E, not charge.
Yes you have to integrate E(y) = Integral( = 0 to )[k*dQ*SIN/r^2] and I believe the result is 2kQ/r^2. This is E(y) at the center for one rod, and of course E(x) = 0 by symmetry, so the total E = 4kQ/r^2 along y.
EDIT. No! I take it all back. Does anyone believe you can double the force on a test charge (i.e., double the field) by spreading the source charge out in a semicircle around the test charge? The correct integration uses Qd/ in place of dQ. The result is E(y) = Integral( = 0 to )[kQ*SIN*d/(r^2)] = 2kQ/(r^2) so total E = 4kQ/(r^2) along y.
The numeric answer = 4*9E9*15E-12/(*0.085^2) = 23.79064 N/C.

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