Calculating Potential Between Conducting Shells

[LocationX]

assuming this thing is ground to the earth, if we placed charge on the inner shell, wouldn't all the charge just travel to the outer shell and into the earth?

 

[gabbagabbahey]

oops, yes you're right!

 

[LocationX]

ok i thought i was missing something there... thanks gabba

 

[LocationX]

My advice is to exploit the spherical symmetry in the problem, by using Gauss' Law to calculate the electric field and then integrating it to find the potential.

one more thing... how do we do this using integration? (the problem with two thin shells)

 

[gabbagabbahey]

Do you mean directly using the integral form of Poisson's equation for a surface charge density (\sigma_i=\frac{Q_i}{4 \pi R_i^2})? Namely:

V(\vec{r})=V_1(\vec{r})+V_2(\vec{r})=\sum_{i=1,2} \frac{1}{4 \pi \epsilon_0} \int_{\mathcal{S}_i} \frac{\sigma_i}{|\vec{r}-\vec{r_i'}|} d^3 r'

 

[LocationX]

I'm not sure if this makes any sense but for the potential to be zero at infinity, don't we need to integrate from infinity to the outer shell for the potential outside the outer shell (which would give 0)

then, to get the potential between R2>r>R1 we integrate from R2 to R1

using gauss law we can find the E-field in between the shells

E=\frac{kQ_1}{r^2}

then the potential would be:

V=-k Q_1 \int_{R_2}^r \frac{1}{r^2} dr

V = kQ_1 \left( \frac{1}{r} - \frac{1}{R_2} \right)

 

[LocationX]

I'm reading in the book that the notation for integration should be:

V=-\int_a^b \vec{E} \cdot d \vec{r}

where the field at a is higher then b

if this is the case shouldn't the integral then be switched
<br /> V=-k Q_1 \int_{r}^{R_2} \frac{1}{r^2} dr <br />

My question how do we know the integral should go from R2 to r and not the other way around?