Electric potential of nested spherical shells

[songoku]

Homework Statement

Two concentric spherical shells have equal but opposite charges. One spherical has radius $a$ and charge $Q$ while the other has radius $b$ and charge $-Q$ where $b > a$. Find $V(r)-V(\infty)$ for the region:
(a) $r>b$
(b) $a<r<b$
(c) $0<r<a$

Relevant Equations

$V=\frac{kQ}{r}$
$V_A - V_B=-\int^{a}_{b} \vec E. d\vec s$
$\int \vec E . d\vec A=\frac{q_{in}}{\epsilon_0}$

For (a), I got $V(r)=0$

For (b), using Gauss law I get the electric field in the region to be $\vec E=\frac{kQ}{r^2}\hat r$, then:

$$V(r)-V(b)=-\int^{r}_{b} \left(\frac{kQ}{r^2}\hat r\right) . (-dr ~\hat r)$$
$$V(r)-0=\int^{r}_{b} \frac{kQ}{r^2} dr$$
$$V(r)=kQ\left(\frac{1}{b}-\frac{1}{r}\right)$$

But if I imagine both charges to be point charges, then:
$$V(r)=V_{\text{by +Q}}+V_{\text{by -Q}}$$
$$=kQ\left(\frac{1}{r}-\frac{1}{b}\right)$$

Where is my mistake?

Thanks

 

[TSny]

For (b), using Gauss law I get the electric field in the region to be $\vec E=\frac{kQ}{r^2}\hat r$, then:

$$V(r)-V(b)=-\int^{r}_{b} \left(\frac{kQ}{r^2}\hat r\right) . (-dr ~\hat r)$$

In the integrand, you should not have the negative sign in $(-dr ~\hat r)$. Instead, $\vec{dr} = dr ~\hat r$. Note that $dr$ is a negative quantity since you are integrating from a larger value of $r$ to a smaller value of $r$.

Considering the direction of the electric field between the shells, do you expect $V(r) - V(b)$ to be positive or negative for $a < r < b$? Does your answer agree with this expectation?

 

[songoku]

In the integrand, you should not have the negative sign in $(-dr ~\hat r)$. Instead, $\vec{dr} = dr ~\hat r$. Note that $dr$ is a negative quantity since you are integrating from a larger value of $r$ to a smaller value of $r$.

Ah I see

Considering the direction of the electric field between the shells, do you expect $V(r) - V(b)$ to be positive or negative for $a < r < b$? Does your answer agree with this expectation?

The direction of electric field in this case is radially outward so I expect $V(r)-V(b)$ to be positive. My first answer does not fit, my second answer fits.

I have another question. How to find potential at $r=a$? I did 2 attempts:
(1)
$$V(a)-V(\infty)=-\int^{a}_{\infty} \frac{kQ}{r^2}dr$$
$$V(a)-0=\frac{kQ}{a}$$
$$V(a)=\frac{kQ}{a}$$

(2)
$$V(a)-V(b)=-\int^{a}_{b} \frac{kQ}{r^2}dr$$
$$V(a)-0=kQ\left(\frac{1}{a}-\frac{1}{b}\right)$$
$$V(a)=kQ\left(\frac{1}{a}-\frac{1}{b}\right)$$

By considering both of them as point charges at the center, my answer matches the second one but I don't really understand why (1) is wrong.

Equation from (1) is like a situation where there is just one charge $Q$. Since point $a$ is also inside another spherical shell so we can't consider the work done to move a charge from infinity to $a$ but need to consider the boundary of the larger shell?

Thanks

 

[TSny]

Ah I see

The direction of electric field in this case is radially outward so I expect $V(r)-V(b)$ to be positive.

Yes.

I have another question. How to find potential at $r=a$? I did 2 attempts:
(1)
$$V(a)-V(\infty)=-\int^{a}_{\infty} \frac{kQ}{r^2}dr$$

Write $$V(a)-V(\infty)=-\int^{a}_{\infty} \vec E_{net} \cdot \vec{dr} = -\int^{b}_{\infty} \vec E_{net} \cdot \vec{dr} -\int^{a}_{b} \vec E_{net} \cdot \vec{dr}$$ Think about what to substitute for $\vec E_{net}$ in each integral on the far right.

 

[songoku]

Thank you very much TSny