# Electric potential between two concentric spherical shells

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1. Jun 4, 2017

1. The problem statement, all variables and given/known data
(The complete problem statement and solution are inside the attached picture)
Two isolated, concentric, conducting spherical shells have radii $R_1=0.500 m$ and $R_2=1.00 m$, uniform charges $q_1=2.00 mC$ and $q_2=1.00 mC$, 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
$$V_f-V_i=-\int_i^f \vec E \cdot d\vec s\,$$
or
$$V=-\int_i^f \vec E \cdot d\vec s\,$$
3. The attempt at a solution
For part (f) and using the results of the previous parts of the problem:
$$V(r)=-\int_{\infty}^{R_2} {E_1}(r) \,dr -\int_{R_2}^r {E_2}(r) \,dr=\frac {q_1 + q_2} {4 \pi \epsilon_0 r}+ \frac {q_1} {4 \pi \epsilon_0 r} - \frac {q_1} {4 \pi \epsilon_0 R_2}$$

And it simplifies to:
$$= \frac {1} {4 \pi \epsilon_0} (\frac {2q_1 + q_2} {r} - \frac {q_1} {R_2})$$
Which is different than what's in the problem's official solution.

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2. Jun 4, 2017

### TSny

For the first term on the far right side, how do you get $r$ to appear in the denominator? Does this term come from the first integral? Note that this integral does not contain $r$ in the limits of integration.