Calculate the electric potential

[Nickclark]

Homework Statement

its not a statement but i was wondering: I'm studying the electric potential and when we have a sphere with inner radious r1 and outer radious r2 that has a charge let's say +Q with a cavity and there is a point charge q inside this cavity so the inner surface will have -q and the outer will have Q+q

Homework Equations

if i want to calculate the electric potential in the cavity and in the sphere between r2 and r1 and outside and on r1 and r2?

The Attempt at a Solution

i think that inside there will be an electric field so V=Kq/r where r is the distance from q
and between r1 and r2 E=0 so the difference must be zero and since there is no charges moving here we will say that through r2>r>r1 the same but how do i calculate it

 

[tiny-tim]

Welcome to PF!

Hi Nickclark! Welcome to PF!

(try using the X₂ icon just above the Reply box )

i think that inside there will be an electric field so V=Kq/r where r is the distance from q
and between r1 and r2 E=0 so the difference must be zero and since there is no charges moving here we will say that through r2>r>r1 the same but how do i calculate it

As you say, inside a conductor E is zero, so the electric potential is constant, and has the value at r₁ that you get from the r ≤ r₁ equation.

 

[Nickclark]

Thanks, but what is its value?
is it V=Kq\r1 through r1<r<r2 or is it V=kq/r2
because if i considered the potential being the integral from a to b for E.dl where b is at infinity so Vb wil be zero then i will get Va:
a) at distance r1 it will be Kq/r1
b) at distance r2 it will be k(q+Q)/r2
but that is wrong because the potential must be constant!
which is why I'm confused

 

[tiny-tim]

ah, good point …

we can add an arbitrary constant to the potential everywhere, so we have to make an arbitrary choice where the zero potential is …

your choice of zero potential at infinity is a lot more sensible than mine!

so yes, do r > r₂ first, then keep the potential constant down to r₁, then lower it in step with the r < r₁ formula

 

[Nickclark]

Oh OK! Got it, Thanks