Question

The figure shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q₁ and radius R; ring 2 has uniform charge q₂ and the same radius R. The rings are separated by a distance 3.00R. The ratio of the electric field magnitudes of Ring 1 and Ring 2 at point P on the common line is 1.59. What is the ratio of charge magnitudes q₁/q₂?

Answer 1

Every point on ring 1 is at distance √(R^{​​​​​​2} + R^{​​​​​​2}) = R√2 away from P.
So if we observe a little bit of charge "q", then it produces field of magnitude
E = kq / (R√2)² = kq / (2R²)

The components of field perpendicular to line joining them cancel out. We're only interested in the horizontal components.

Each little bit of charge on ring 1 produces a field E that points at 45° from the horizontal.
The purely horizontal component of that field is E*cos45° or using similar triangles E * (√2 / 2)
= kq/R^{​​​​​​2} * (√2 / 4)

Since it is the same for every point on the ring, add that up around the ring and you get magnitude of field produced by the entire ring.
E^{​​​​​​}_{​​​​​​1} = k (q₁)/R^{​​​​​​2} * (√2 / 4)

Repeat same for ring 2.,
Field produced by a little bit of charge is E = kq / [√(R^{​​​​​​2} + (2R)²) ]²
E = kq / (5R²)

Using similar triangles, horizontal component =

E * (2R) / {√[R^{​​​​​​2} + (2R)²]}
= E * 2 / √5
= kq / (5R²) * (2 / √5)
= kq/R^{​​​​​​2} * 2/ (5√5)

So, magnitude of total field produced by ring 2 is :
E_{​​​​​​2} = k (q₂)/R^{​​​​​​2} * 2/(5√5)

We know,

E_{​​​​​​1} = 1.59 E_{​​​​​​2}
Substitute and simplify,

q_{​​​​​​1} * √2 / 4 = 1.59 q_{​​​​​​2} * 2/(5√5)

q_{​​​​​​1} / q_{​​​​​​2} ​= 0.804 ~ (0.80 approx) _ Ans