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

A shell is an "empty" sphere, all the charges are located on its surface.
Gauss law relates the electric flux passing through chosen closed surface with the total amount of charge enclosed by that surface.
Just chose another shell whose r < R of the shell given in your original task and this should be trivial. Hint: you're not enclosing any charge.

Say if you're not allowed to use gauss law, mention what approach should I take.

Answer 2

yes we can apply gauss's law

Answer 3

E inside a spherical shell is 0

Answer 4

ok... can you finish or is something confusing you still?

Answer 5

Then V will be constant?

Answer 6

will you be able to give me a proof here? It will be better

Answer 7

So what you get is that the electric field is zero. Gauss law:
$$\Phi = \oint \vec E\cdot d\vec A = \frac{Q_{\text{encircled}}}{\epsilon_0} = 0$$
by definition electric flux is:
$$d\Phi = \vec E \cdot d\vec A $$
which means that the electric field is 0 as you managed to conclude nicely. While at the same time the potential is just the difference between the electric fields in two different points:
$$dV = -\vec E \cdot d\vec s \\
V = -\int_S\vec E \cdot d\vec s$$
Point being that this is a path integral that evaluates E in some two points even if E is a constant different than 0 this integral would not evaluate to anything except 0.

Answer 8

Maybe you're confused about the resolution of the path integral? It's not like an indefinite integral where the result of the integration is an infinite family of functions different by a constant, you know what you get. Definitive integrals don't have ambiguities in them.
$$V = -\int_S\vec E \cdot d\vec s = -\int_{r_0}^{r_1}Eds = -E\int_{r_0}^{r_1}ds=-Es|_{r_0}^{r_1} = -E(r_1-r_0)$$

but E = 0

Answer 9

and in retrospect I take back what I said about E being a constant and voltage turning out zero. That's not true I can see that now. Otherwise how would capacitors work.

Answer 10

Yeah .... V= constant and V=0 may mean the same in this case, E=0 and E=-grad V. so grad V=0 implies V=0.. may be constant V is considered 0

Answer 11

Ah, well now I get what might be a bit confusing.
Yes, if you look at electric potential (that is also sometimes marked with V) as a function then it is a constant function all the way inside the shell. In fact it would be the value of the potential on the surfae of the sphere.
But if you take V to mean "voltage" (which is usually marked with V as well) then we're talking about the differences of electric potentials in two points.

Answer 12

Since your first question referred to finding potential I just assumed the second V is meant to be voltage.

Answer 13

no.. V is electric potential, its not the same as volt, which is measure of potential difference

Answer 14

Volt is a unit in which you measure voltage. Electic potential is measured in Joules and is more often than not marked with the letter U because this ambiguity between the V and U for voltage and electric potential has been around for a while.

You know electric potential inside a metal shell isn't zero because it's among the first tasks people solve in EDM, or at least you should be able to google it.

Answer 15

So the question under consideration, the second part is wrong?

Answer 16

Well if it means anything except voltage it would be.

Answer 17

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html

quick google shows that.

Answer 18

I really hope I'm not wrong here, that would be so embarrassing :D

Answer 19

so it will be constant i guess

Answer 20

i've posted a response on the maths forum which is on all fours with @ljetibo

just adjust the integration intervals you used for part 1 and you should be good