Wow, I cannot believe I missed that.
So the revised eq would be:
$$\frac{360}{150} = \frac{.062}{.19} * (\frac{(.19^2+a^2)}{(.062^2+a^2)})^\frac{3}{2}$$
$$(\frac{360}{150} * .\frac{.062}{.19})^\frac{2}{3} = \frac{(.19^2+a^2)}{(.062^2+a^2)}$$
$$(\frac{360}{150} *...
Homework Statement
The electric field on the axis of a uniformly charged ring has magnitude 360 kN/C at a point 6.2 cm from the ring center. The magnitude 19 cm from the center is 150 kN/C ; in both cases the field points away from the ring.
Homework Equations
1. Find the ring's radius.
2...