Two 10-cm-diameter charged rings face each other, 15 cm apart. the left ring is charged to − 20 nc and the right ring is charged to 20 nc. what is the magnitude of the force on a proton at the midpoint

Radius of the rings = 10 / 2 = 5 cm 0.05m
Distance between them D = 15 cm = 0.15 m
The charge of left ring is Ql = -20 x 10^-9 C
The charge of right ring is Qr = 20 x 10^-9 C
Since it the electric field at the midpoint, it is same for both rings and the there are opposite charges the electric filed will be sum of electric field of both charges.
Electric field on the left El = (k x Ql) / (D/2)^2 = (4 x k x Ql) / D^2
Electric field on the Right Er = (k x Qr) / (D/2)^2 = (4 x k x Qr) / D^2
Electric field E = El + Er =  ((4 x k x Ql) / D^2) + ((4 x k x Qr) / D^2)
E = 4k/ D^2 (Ql + Qr); k is Coulombs constant = 8.99 x 10^9 Nm2Câ’2
E = 4 x 8.99 x 10^9/ 0.15^2 (2n + 2n) = 1598 x 10^9 x 4 x 10^-9 = 6393 N/c
Force f = E x q, q being the charge at the mid point
Force = q x 6393 N/c = 6.393q x 10^3 N