Electrostatics
A uniformly charged ring of radius R has a total charge Q.
1) What is the magnitude and direction of the electric field along the axis of the ring?
2) What is electric field potential along the axis of the ring?
3) What is the electric field at the center of the ring?
4) What is the electric potential at the center of the ring?
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Solution Preview
1) Charge density on the ring = Total charge/Circumference = ρ = Q/2ΠR
Let us consider a small element dl of the ring. Charge dq on this element is given by :
dq = (Q/2ΠR) dl
Electric field intensity due to the element dl at a point P on the axis of the ring
= dE = (1/4Πε0)dq/L2 = (1/4Π L2ε0)(Q/2ΠR) dl
dE can be resolved into components as shown in the fig. If we consider an element diametrically opposite
to the first one, the electric field due to the same will be equal in magnitude but opposite in direction. The
same can also be resolved into ...
Solution Summary
This step-by-step solution explains how to find the electric field of a uniformly charged ring.