Question
See the figure (http://s19.postimage.org/fugnancyr/Exam2_No1.png). A circular ring of charge (total charge = Q and radius = a) is located on the xy-plane centered at the origin.
(A) Determine the electric potential at a point (0, 0, r), i.e., on the z-axis at z = r.
(B) Determine the electric potential at a point (r, 0, 0), i.e., on the x-axis at x = r. Leave the answer in terms of an integral, but simplify the integral so that you can calculate its numerical values.
(C) Make a graph that shows both V(0,0,r)/V(0,0,0) and V(r,0,0)/V(0,0,0) versus r. Use graph paper to plot the points accurately, by hand, for a = 1.0 m and 0 ? r ? 3 m. You will need to use a computer to calculate the potential on the x-axis. E.g., you could use Wolfram Alpha or any other software.
Explanation / Answer
p- charge density q - charge A - area ke - Coulomb constant (A) You know that p=q/A p = q / (pi*r²) q = pi*p*r² dq = pi*p*r*dr (r is the radius) Now, you also know that E is equal to: E = ke ? dq/d² , where d is the distance. If you look closely you'll see that d² is equal to z² + r², therefore: E = ke ? dq / (z² + r²) E = pi*p ? r / (z²+r²) dr Now, this will compute the electric field at any point. However we can see by symetry that the electric field will only have a z component, because the x component will cancel each other. Therefore we have to add: E = pi*p*ke ? r * cos(tetha) / (z²+r²) dr theta is the angle that the electric field does with the z-axis. Again, as you probably remember cos(tetha) will be the quoficient of the adjacent side by the hypotenuse, or in this case, z / hypotuse. And the hypotenuse is equal to the sqrt(x²+r²), therefore we conclude that cos(theta) = z / sqrt(x²+r²) Substituting: E = pi*p*ke ? r * z / (x² + r²)^3/2 dr If you integrate this between 0 and r (radius) you'll get: E = 2*pi*p*ke [ 1 - x / sqrt(x²+r²) ] (B) By symmetry you can see that the electric field will only have a x component. All you have to do is compute the electric field as if the source was just a line of length l = 2r. Since you know the density you know that: q = p * l q = 2 * p * r dq = 2 * p * dx Integrate it: E = 2*ke*p ? 1/x² dx (C) Kind of hard to explain it by text, so I'll leave that one for you :P Good luck!