# Electric Field & Potential from Two Conductors

1. Dec 7, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Two isolated, concentric, conducting spherical shells have radii R1=0.500 m and R2=1.00 m, uniform charges q1=2.00 mC and q2=1.00 mC, and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a) r=4.00 m, (b) r=0.700 m, and (c) r=0.200 m? With V=0 at infinity, what is V at (d) r=4.00 m, (e) r=1.00 m, (f) r=0.700 m, (g) r=0.500 m,(h) r=0.200 m, and (i) r=0? ( j) Sketch E(r) and V(r).

2. Relevant equations
$\oint \vec{E}\cdot\vec{dA}=\frac{q_{encl}}{\varepsilon_0}$
$V_f-V_i=-\int_i^f\vec{E}\cdot d\vec{s}$
3. The attempt at a solution
I'm not having any trouble with the electric fields so I will just list the results:

$r<R_1, \hspace{5mm} E=0$
$R_1<r<R_2, \hspace{5mm} E=\frac{q_1}{4\pi\varepsilon_0 r^2}$
$R_1<R_2<r, \hspace{5mm} E=\frac{q_1+q_2}{4\pi\varepsilon_0 r^2}$

I am quite confused by the electric potential however, taking $V_i=0$ at infinity, I don't understand the process of calculating the electric potential.

$$V=-\int_{\infty}^{r}\vec{E}\cdot d\vec{s}$$

I'm not sure what to replace $d\vec{s}$ with, perhaps a vector $d\vec{r}$ that goes outwards radially and then integrating along the radius. Although, suppose I want to integrate to an $r>R_2$, the Electric Field changes throughout the the integration at points $r=R_1$ and $r=R_2$. How would I go about doing this?

Edit: Solved, split integral up so that the correct electric field is in place for the desired radius

Last edited: Dec 7, 2015
2. Dec 8, 2015

Well done.