# What is the charge of a point at the center of a ring

1. Mar 7, 2012

### nateja

1. The problem statement, all variables and given/known data
In the figure (I'll try to find it) a ring of radius .71m carries a charge +580nC uniformly distributed over it. A point charge Q is placed at the center of the ring. The electric field is equal to zero at field point P, which is on the axis of the ring, and 0.73 m from its center.The point charge Q is... then give's multiple choice solutions.

2. Relevant equations

E = 1/(4*pi*epsilon_0) Q/r^2

3. The attempt at a solution
My first attempt got me the wrong value. I tried to find the E-field and integrate the little bits of charge around the disk but my answer was too large. I know that the E field from the ring will push put an E field on P that goes out in the x-direction along the axis. So in order for the E field to be 0 at point P, the charge Q must put out an E field of equal magnitude but opposite direction.

Using the below formula, I then set the E equation = to the value I calculated and solved for Q, but then got the answer that was too large.
E = k*(λ*.73)/(.73^2+.71^2)^(3/2)*2*pi*.71 (got this equation from integrating and using λ = 580nC/m

My friend tried a different way: plugging in 580nC for the charge (where I had used it for the charge density) and (.73)^2+(.71)^2 for r^2 into. We both got an E = 5029 N/C. Then we set that value equal to the E equation:

-5029 = k*q/r^2
q = (-5029/k)*r^2 = (-5029/4*pi*8.85*10^-12)*(.73^2) = -2.97*10^-7 (-290nC)

The answer is supposed to -210nC

2. Mar 7, 2012

### Staff: Mentor

The charge given for the ring is the total charge, not the charge density.