Question

Two 10-cm-diameter charged rings face each other, 20.0 cmapart. Both rings are charged to + 10.0 nC . What is the electric field strength

A.)at the midpoint between the two rings?

B.)at the center of the left ring?

Answer 1

a) Zero.

If we put a test charge there, the force from the left ring exactly cancels the force from the right ring - due to the symmetry.

b) The left ring itself has no contribution to the total field at its center, same argument as part a).

The entire field is from the right hand ring.

Imagine the axes of the rings are horizontal. Call the center of the left ring 'P'.

Each element (small piece) of the right ring is a distance d from P where

d = sqrt(20^2+5^2) = 20.6cm = 0.206m

The angle subtended by the element at P = tan-1(5/20) = 14deg

If each element carries charge q then the field from it at P is kq/(0.206)^2 (where k=Coulomb's constant, 9x10^9units)

The horizontal component of this field contribution at P is kqcos14/(0.206)^2

Not that all the field components perpendicular to the axis will cancel (symmetry again). We just have to sum the horizontal (axial) field components.

The total field = the sum of the fields from all the elements. If the total charge is Q, this is just

kQcos14/(0.206)^2

= (9*10^9)(10*10^-9)cos14/(0.206)^2

= 2058 N/C