Homework Help: Field at a distance caused by a ring charge

1. Jul 10, 2017

chopnhack

1. The problem statement, all variables and given/known data
2.75 mC is uniformly distributed on a ring of radius 8.5 cm. Find the electric field on the axis at (a) 1.2
cm, (b) 3.6 cm and (c) 4.0 m from the center of the ring.

2. Relevant equations

E = (kQx)/(a^2+x^2)^(3/2)

3. The attempt at a solution

This question really boils down to a calculator issue, operator error issue or the abyss of unknown. When I perform the calculation on two different calculators using different methods, including a graphing calculator that has memory allowing me to repeat an equation previously used and change its variables, I get a strange answer for the second part 3.6cm. It ends up larger by one order! It can't be because its further from the charge, but each time I calculate it I get an erroneous value.

a) 4.695 x 10^5 N/C
b) 1.13 x 10^6 N/C
c) 1.54 x 10^3 N/C

Please don't say throw out the calculator because I checked it with a 30 year old Sharp 531D and I get the same results!! I must be missing something. Anyone able to assist? Most appreciated, thanks!

2. Jul 10, 2017

TSny

Are you sure it can't be? What is the field at x = 0?

3. Jul 10, 2017

chopnhack

Fair point, the value at x = 0 is zero, but I think that is almost like the argument of no charge inside a hollow sphere. In this case, the center is the ring, it being just a slice of a sphere.

I am not sure it can't be, but it doesn't seem logical in the sense that as you get further from the field you should see a diminished field, as is the case for all other charges we have dealt with so far.

Have you tried the calculation?

4. Jul 10, 2017

TSny

Only the x-component of the field survives. Suppose you had a single point charge located on the y axis at say, y = .30 m. Now consider different points on the x axis, say x = 0, x = .10 m, x = .20 m, ..., x = 1.0 m. Can you see how the x-component of the field of the point charge varies for these points?

Yes, I get your results. Since you have a graphing calculator, you might try plotting your function E(x).

5. Jul 10, 2017

chopnhack

Yes, the opposing sine theta portions cancel each other leaving just the cosine portion.
I will give that a go and see if I can get it to display a graph.

That other bit you mentioned on a point charge at y - I can somewhat see what you mean, due to the circular nature of the ring, but I want to work it out later when I have some time.