Question

Three narrow concentric rings of radii 50 mm, 70 mm, and
90 mm are centered about the origin, with the axis of symmetry of each ring oriented along the y axis. The charge on the inner ring is 1.0 mC, that on the middle ring is -2.0 mC, and that on the outer ring is 1.0 mC.
(a) What is the electric field magnitude at (0, 100 mm, 0)?
(b) What quantity of charge must be placed at (0, -100 mm, 0) to make the electric field magnitude zero at the position calculated in part a?

Answer 1

The electric field of a ring within the plane in which the ring is placed, or at $z=0$ in this problem, can't be calculated with some easy expression, since calculations will lead us to elliptical integrals. However, it is possible to get an expression for a field above a charged ring with radius $r$ and total charge $Q$:

$$E=k\frac{yQ}{(y^2+r^2)^{3/2}}.$$

Calculate the field at $(0,100,0)$ mm for each ring, inner, middle, outer:

$$E_I=643.1\text{ MV/m},$$

$$E_M=-988.3\text{ MV/m},$$

$$E_O=369.1\text{ MV/m},$$

therefore according to the superposition principle, at the height of 100 mm the total field is

$$E=E_I+E_M+E_O=23.9\text{ MV/m}.$$

To get zero field at that point, we need to place a charge 100 mm below the center of the rings which will provide a field equal to that of the rings:

$$E=k\frac{q}{R^2},$$

$$q=\frac{ER^2}{k}=\frac{23.9\cdot10^6\cdot(0.1-(-0.1))^2}{9\cdot10^9}=1.06\cdot10^{-4}\text{ C}.$$

The charge should be positive.