Consider a small segment of the ring that subtends an angle d at the center and its azimuthal angle (angle measured from a fixed diameter in this case) be As q = q0sin dq/dl = q0cos(d/dl) = q0cos(d/rd) = (q0cos)/r Now charge on the element considered, dq = (dq/dl) rd = [(q0cos)/r](rd) = q0cos d Now, you can check that due to the charge distribution given, Half of the ring is positive and the other half is ve. So for an element such as ours, the component of its electric filed at the center, along the azimuthal line is canceled out due to similar element on the following quarter circle. Also, it is reinforced by similar tqo -ve elements from the opposite side E = 2(1/4)0pi((dq)/r2)cos = (2q0/4r2)0picos2d Evaluate the integral and find E
Moderation Team
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where 
at the center and its azimuthal angle (angle measured from a fixed diameter in this case) be 
)0
pi((dq)/r2)cos
picos2
