# Potential Difference b/w concentric shells -- confusion

## Homework Statement

:[/B]

This is isn't exactly a problem,but actually something i don't understand in the book i was following, so there this art. about 'Principle of a Generator', whose description is given as, " A generator is an instrument for producing high voltages in the MeV range.
Its design based on the principle that if a charged conductor (labeled A) is brought in contact with a hollow conductor (labeled B), all of its charge transfer to the hollow conductor no matter how high the potential of the later may be.
"

## Homework Equations

I have added the supporting equations as attachments (images ).
Img1:
Img2:
img3
img4

3. The confusion :
(check all images, mainly the 3 equations in them)

In those attached images, i think the the potential at shell A(1st equation i.e Va =) and B (second equation Vb =) must be the sum of potential due to shell A and B at the required shell (the shell for which we want to find the potential for/at ), so instead of taking difference while calculating the potential at shell A and B, we should simply add them, no ? (Assuming charges are positive or keeping them as variables, there really isn't any reason to assume otherwise ?)
But author has, subtracted charges i.e qa/ra -qb/rb.
I think we should have added them. (i.e qa/ra + qb/rb. )

I hope i'm not too confusing, i'm newbie and don't use internet much.

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kuruman
Homework Helper
Gold Member
The principle is quite simple and has to do with the fact that a conductor is an equipotential. If you place charge on a conductor, it all goes to the surface. This follows from Gauss's Law and the fact that the electric field inside a conductor is zero. When you have a charged conductor inside the cavity of another conductor and you provide a conducting path to the wall of the cavity, what do you get? Answer: A single conductor with a (differently shaped) cavity. All the charge on the inner conductor will go to the surface of the outer conductor otherwise the new composite conductor will not be an equipotential.

IonizingJai
The principle is quite simple and has to do with the fact that a conductor is an equipotential. If you place charge on a conductor, it all goes to the surface. This follows from Gauss's Law and the fact that the electric field inside a conductor is zero. When you have a charged conductor inside the cavity of another conductor and you provide a conducting path to the wall of the cavity, what do you get? Answer: A single conductor with a (differently shaped) cavity. All the charge on the inner conductor will go to the surface of the outer conductor otherwise the new composite conductor will not be an equipotential.
I understand that, but that is not what i'm confused about, i'm confused about what i have mentioned in my OP, i.e In those attached images, i think the the potential at shell A(1st equation i.e Va =) and B (second equation Vb =) must be the sum of potential due to shell A and B at the required shell (the shell for which we want to find the potential for/at ), so instead of taking difference while calculating the potential at shell A and B, we should simply add them, no ? (Assuming charges are positive or keeping them as variables, there really isn't any reason to assume otherwise ?)
But author has, subtracted charges i.e qa/ra -qb/rb.
I think we should have added them. (i.e qa/ra + qb/rb. )
I want to know what the author has written in the book is right (probably) and understand how ?

kuruman
Homework Helper
Gold Member
OK, sorry I misunderstood what you were asking. If the outer shell has charge ##q_B## and the inner sphere has charge ##q_A##, then by Gauss's Law the electric field in region I (##r>r_B##) is ##E_I=k(q_A+q_B)/r^2##, which makes the potential ##V_I=k(q_A+q_B)/r##. Thus, the potential of the outer shell is $$V_B=k\frac{(q_A+q_B)}{r_B}.$$In region II ##(r_A<r<r_B)## the electric field is ##E_{II}=kq_A/r^2##. The potential is ##V_{II}=kq_A/r+C##, where ##C## is determined by making sure that the outer shell is an equipotential, namely ##V_I(r_B)=V_B.## Thus, $$k\frac{q_A}{r_B}+C=k\frac{(q_A+q_B)}{r_B}~\rightarrow C=k\frac{q_B}{r_B}$$
Therefore, $$V_{II}=k \left(\frac{q_A}{r}+\frac{q_B}{r_B} \right)~\rightarrow~V_A=k \left(\frac{q_A}{r_A}+\frac{q_B}{r_B} \right)$$which makes$$V_A-V_B=k \left(\frac{q_A}{r_A}-\frac{q_A}{r_B} \right)$$So the bottom line is the same, but not how one gets to it.

IonizingJai
That's quite right !
i think i was way too lost in wondering why the author in the book wrote the equations the way he did, and i should have tried your line of approach(which i initially was saying saying should be correct) and didn't proceed further to subtract the equations of Va-Vb.

Thanks for the help !