Calculating the Electric Potential from an Electric Field problem

AI Thread Summary
To calculate the electric potential from an electric field for a nonconducting sphere with a radius of 2.31 cm and a charge of +3.50 fC, the potential at the center is set to zero. For a radial distance of 1.45 cm, the electric field E is calculated using the formula E = kqr/R^3, while for r = R, E is determined by E = kq/r^2. The initial calculations yield -5.37 x 10^-4 V for r = 1.45 cm and -0.001 V for r = R. Confusion arises regarding the integration variable ds, with the manual suggesting ds = r/2 instead of r. Clarification is sought on the integration limits and the correct formula for V after integration.
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Homework Statement



A nonconducting sphere has radius R = 2.31cm and uniformly distributed charge q = +3.50 fC. Take the electric potential at the sphere's center to be Vo = 0. What is V at radial distance (a) r = 1.45cm and (b) r = R?

Homework Equations



V = the negative integral from i to f of E * ds

For a, E is kqr/R^3.

For b, E is kq/r^2

The Attempt at a Solution



When I plug all this in I get -5.37 x 10^-4 V and -.001V, respectively. However, when I look at the answer explanation in the student manual, it seems to indicate that ds (or in this case rs) = r/2. Why is it r/2? How is the book getting that for ds? At first I thought ds was equal to r, but I guess I'm not grasping something. Could someone explain why this is so? I basically have the right answer, but only if I divide both my answers by 2.
 
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what are your limits of integration and what's your formula for V after you integrate?
 

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