Question

Need help with part ii).
i)Find the potential inside and outside a spherical shell of radius R which carries a uniform surface charge. Set the reference point at infinity.
ii)If the reference point was at somewhere inside the spherical shell, show that V=0.

Answer 1

@ganeshie8

Answer 2

if i understand this, you are now looking at the work required to move a unit charge from somewhere inside the sphere to wherever.

the potential will be zero inside as the field is zero and \(E_r = - \dfrac{dV}{dr} \implies V = const \) and you're saying that somewhere inside is zero potential so it's all zero

but potential should be negative outside the sphere as a unit positive charge would fly off to infinity if placed just outside the sphere ..... thereby losing potential energy

reckon easiest thing is to reverse the integral intervals on the work you have already done and make sure the signs comform to the physics

Answer 3

\[V(r)=-\int\limits_{O}^{r} E. dl\]

Answer 4

in the first case

Answer 5

@IrishBoy123

Answer 6

So we put O=infinity

Answer 7

In the second case, O will be somewhere inside the spherical shell

Answer 8

inside, i think it would be this

\(V = - \int_{r_o}^R \vec E \bullet d \vec r = - \int_{r_o}^R 0 dr = 0\)

where \(r_o\) is wherever you set V = 0 inside the shell

for outside, i think this

\(V = - \int_{r_o}^R \vec E \bullet d \vec r - \int_R^r \vec E \bullet d \vec r\)

\(= - \int_{r_o}^R 0 dr - \int_R^r \dfrac{kq}{r^2} dr\)

\(= kq \left( \dfrac{1}{r}- \dfrac{1}{R}\right)\) which is going to be negative as \(r > R\)

so it matches the physical idea that you lose potential as you move away

does that look like what you did in the first place?

Answer 9

Thanks... :)... @IrishBoy123

Answer 10

if @ljetibo is happy with that, i'd go with it

Answer 11

fine by me, I think I've made that same integral in fact. (the first one)