The figure shows positive point charges $q_1$ and $q_2$ on a ring of radius $R.$ Charge $q_1$ is on the $x$ axis a distance $R$ to the left of the origin. Charge $q_2$ is placed somewhere on a circle of radius $R$ centred at the origin, at a location that makes an angle $$ with the positive $x$ axis, as shown. The electric field strength $E$ is measured at the origin.

$($a$)$ Show that the angle $$ satisfies

$cos=\frac{q_1^2+q_2^2-E^2(4lo{n}_0{R}^2{)}^2}{2q_1{q}_2}.$

$($b$)$ Find $$ in degrees for

$q_1=2\mathrm{.}0nC,q_2=3\mathrm{.}0nC,E=1\mathrm{.}0{10}^{6}N{C}^{-1},R=5\mathrm{.}0{10}^{-3}m.$

Note that $$ can be positive or negative and still yield the same electric field strength at the origin, so just write down the positive value.