18. We use Eq. 23-3, assuming both charges are positive.
Eleft ring =Eright ring evaluated at P
q1R
4πε0(R2+R2)
3/2=q2(2R)
4πε0((2R)2+R2)
3/2
Simplifying, we obtain
q1
q2
=22
53/2
≈0.51 .
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Problem 18 on the Electric Fields with Solution, Exercises of Electrical Engineering
This is solution to problems related Electrical Engineering course. It was suggested by Prof. Bhooshan Sawhney at Shree Ram Swarup College of Engineering and Management. It includes: Electric, Field, Surface, Center, Magnitude, Radius, Normal, Magnetic, Dpole, Moment
Typology: Exercises
2011/2012
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- We use Eq. 23-3, assuming both charges are positive.
Eleft ring = Eright ring evaluated at P q 1 R 4 πε 0 (R^2 + R^2 )^3 /^2
q 2 (2R) 4 πε 0 ((2R)^2 + R^2 )^3 /^2
Simplifying, we obtain q 1 q 2