Consider a thin ring of insulative material with a radius of 5 meters where the left half of the ring has been given a net charge of +70 nC and the right half has a net charge of -70 nC, distributed evenly along each half of the ring. If a +1 nC test charge is placed in the center of the ring, what is the electric force on it in both magnitude and direction?

Question

Consider a thin ring of insulative material with a radius of 5 meters where the left half of the ring has been given a net charge of +70 nC and the right half has a net charge of -70 nC, distributed evenly along each half of the ring.  If a +1 nC test charge is placed in the center of the ring, what is the electric force on it in both magnitude and direction?

Anonymous · Answer

both resultants pull or push to the right so just do the left half ring and double the result. In fact just do half of the left ring and quadruple the result :) All we need is the x component of force, we know the up and down cancels. so find q = charge in coulombs / angle dT in radians = total charge on the left / pi radians = 70 nC / 3.14159 = 22.3 nC/radian so x component of force on Q which we can call E because one coulomb is at the center d Ex =k * 22.3 nC * d theta cos theta / R^2 Ex = (22.3 nC * k / 25 ) integral 0 to pi/2 of cos Theta d Theta = (22.3 nC * k / 25 ) * (sin pi/2 - sin 0) well the integral seems to be one :) now multiply by four because we only did a quarter of the circle :)