Recent content by SlyCedix

800 nC

That was it, thanks, answer was 8.8cm

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...