Potential Difference b/w concentric shells -- confusion

  • #1

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:
https://drive.google.com/file/d/1_oNyWdSR2YDWiRMAoX8NY6OKrQJu5sEQ/view?usp=sharinghttps://drive.google.com/file/d/1_oNyWdSR2YDWiRMAoX8NY6OKrQJu5sEQ/view?usp=sharing
Img2:
https://drive.google.com/file/d/1DbYC36dxTOJjGogU80YJgDt10EBlhRn8/view?usp=sharing
img3
https://drive.google.com/file/d/1hh0D-5kN1r2iy6BZ_s_-3Z-625aTuz8C/view?usp=sharing
img4
https://drive.google.com/file/d/1HrBINrhGZynXOi3McwwLtba2rfjdgbSs/view?usp=sharing

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.
 

Answers and Replies

  • #2
kuruman
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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.
 
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  • #3
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 ?
Anyway, thanks for replying.
 
  • #4
kuruman
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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.
 
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  • #5
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 !
 

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