Isolated and non isolated conductors

[amjad-sh]

I picked these questions below from a chunk of exams most of them are solved except this one.
Questions 2 and 5 are confusing me.
In question 2 s1 is supposed to be isolated and in 5 its not .
if anybody can explain to me briefly what is happening or can solve them to me he will be so appreciated.
thanks

 

[collinsmark]

I picked these questions below from a chunk of exams most of them are solved except this one.
Questions 2 and 5 are confusing me.
In question 2 s1 is supposed to be isolated and in 5 its not .
if anybody can explain to me briefly what is happening or can solve them to me he will be so appreciated.
thanks View attachment 71539

Question 2:

The shell (the question calls it a "hollow sphere," I'm calling it a "shell" -- same thing), S₂ is isolated and has no net electric charge on it (it is initially neutral).

a- Determine the charge distribution one the sphere, S₁ and also the shell S₂. Hint: Use Gauss' law for this. Also note that the electric field within a conducting material is always zero (in the material itself -- this doesn't necessarily apply to hollow portions [i.e. cavities] inside the material, but rather just inside the material itself). [Edit: it doesn't necessarily apply to the surface either. You know that the shell S₂ has no net charge, but it might very well have equal but opposite charges on its inner and outer surfaces.]

b- Determine the new potential of the inner sphere. Hint: you'll need to do some integration as part of the solution. How to you calculate the potential from a given electric field? Assume the potential at infinity is zero.

c- Same as part b, but you only need to integrate to the shell, S₂.

Question 5:

a- The potential of the inner sphere S₁ is brought back to potential V₁. The shell S₂ is has zero net charge. Find the charge on S₁.

b- Same situation as above, but now the shell S₂ is brought to a potential of zero. Find the charge on S₁.

 

[amjad-sh]

Lets begin from question 1 :
my solution :
1) V∞ -V1=₀∫^∞ E.dr
what I got is V1=Q1/4∏εR1.
then Q1=4∏εR1*V1
(but since in question 5 he said that s1 is no more isolated rather maintained at potential V1 then I'm not sure about this solution).2) a) from gauss law E*₀∫^rds=Qin/ε (where r>R2)
then Qin=Q1+Q' and as E must be zero inside the "shell" then Qin=0 and then Q'=-Q1.(Q' is the charge on the inner surface of the shell and I chose spherical gussian surface surrounding sphere of radius R2 consequently charge on the surface of the shell will be +Q1 )

b) V∞-V1'=-_R1∫^∞E.dr
we get that V1=Q1/4∏εR1.

c) V2-V1'=-_R1∫^R2E.dr
we get that V2=Q1/4∏εR2.

3) a) if V2=0 then charges on the surface of the shell will be zero then Q2 =0.
b) V"1=_R1∫^R2E.dr
V"1=(Q1/4∏ε)*(1/R2-1/R1).

Lets skip Question four.
In question 5 I didn't know what to do.
I'm not sure about the solutions I wrote above so if you can help please and thanks btw.

 

[collinsmark]

Lets begin from question 1 :
my solution :
1) V∞ -V1=₀∫^∞ E.dr
what I got is V1=Q1/4∏εR1.
then Q1=4∏εR1*V1
(but since in question 5 he said that s1 is no more isolated rather maintained at potential V1 then I'm not sure about this solution).

That looks about right to me so far.

2) a) from gauss law E*₀∫^rds=Qin/ε (where r>R2)
then Qin=Q1+Q' and as E must be zero inside the "shell" then Qin=0 and then Q'=-Q1.(Q' is the charge on the inner surface of the shell and I chose spherical gussian surface surrounding sphere of radius R2 consequently charge on the surface of the shell will be +Q1 )

If I understand, yes you are correct. Your previously calculated value of Q₁ is on the outside of the central sphere. -Q₁ exists on the inside of the S₂ shell and +Q₁ exists on the outside of the S₂ shell.

b) V∞-V1'=-_R1∫^∞E.dr
we get that V1=Q1/4∏εR1.

I think you are missing a few steps here. You need to break the integral up into three parts, since there are three different regions of electric field.
Infinity to R₃
R₃ to R₂
and
R₂ to R₁

You can then sum the individual results together to find the potential from infinity to R₁.

c) V2-V1'=-_R1∫^R2E.dr
we get that V2=Q1/4∏εR2.

I interpret the problem as asking you to find the potential between R₂ and infinity, not R₂ and R₁.

Your answer seems correct though. Just realize that that is the potential difference of S₂ from infinity, i.e., it is not V₂ - V'₁.

(In other words, it's V₂ - 0.)

3) a) if V2=0 then charges on the surface of the shell will be zero then Q2 =0.

That's not right. But you can fix it easily enough.

What it means to Earth shell S₂ is the electric field from infinity to R₃ must be equal to zero. Given that the sphere S₁ has a Q₁ charge on it, what does Gauss' law say about what the total charge must be on S₂ (where Q₂ is the net charge on shell S₂)?

Edit: Let me rephrase that for clarity. Shell S₂ is earthed. That means the electric field for all space outside the shell S₂ is zero.

• If so, what does Gauss' law say about the total charge of the S₁ and S₂ combination.?
• Since you know that S₁ has charge on it, what charge must S₂ have on it?

b) V"1=_R1∫^R2E.dr
V"1=(Q1/4∏ε)*(1/R2-1/R1).

Lets skip Question four.
In question 5 I didn't know what to do.
I'm not sure about the solutions I wrote above so if you can help please and thanks btw.

Part a). You don't know what the charge is on S₁, but you do know it's potential (it's potential is V₁). You don't know the value of this charge, but you can still give it a name for now. Let's just call it q for now (or use whichever name you prefer). Set up the charge distribution on shell S₂ in terms of the variable q. Now breakup the integral and solve three parts of the integrals. (don't forget about the negative involved $V = - \int_\infty^R \vec E \cdot \vec{d \ell}$.)

You know that the sum of all the individual sections must be V₁. With that, solve for q.

Part b This time, the shell S₂ is earthed (i.e., grounded, has a potential of 0 Volts). Essentially that means that this time you only need to integrate from R₂ to R₁ instead of from all the way from infinity to R₁.

 

[amjad-sh]

OK thanks,I'll show my corrections:
In question 2) b)V∞-V1'=-(_R1∫^R2Edr +_R2∫^R3Edr +_R3∫^∞Edr)
we will get that V1'=Q1/4∏εR1 -Q1/4∏εR2+Q1/4∏εR3

c)V∞-V2=-(_R2∫^R3Edr(0)+_R3∫^∞Edr)
then we will get that V2=Q1/4∏εR3.

In question 3) a) Electric field outside shell is zero,lets apply gauss law E∫ds=Qin/ε then
E4∏r²=Qinc/ε.(r>R3)
then Qinc=0 and as Qin=Q1+Q2 then Q2=-Q1.(But i think charges on the outer surface of the shell will be zero and -Q2 whice is -Q1 will surround the inner surface the inner surface of the shell).

b)v∞-v1"=-_R1∫^∞Edr
thus -v1"=-(_R1∫^R2Edr +_R2∫^R3Edr (whice is zero since its Ein=0 ) +_R3∫^∞Edr(whice is zero).
we will get that V1"=Q1/4∏ε(1/R1-1/R2).5) a) Let q be the charge of s1, the charge distribution on the conductors will be same as previous parts.
V∞-V1=_R1 ∫^R2Edr+_R2∫^R3Edr(0)+_R3∫^∞Edr
then V1=q/4∏εR1 -q/4∏εR2 +q/4∏εR3
Q1/4∏εR1=q/4∏ε(1/R1-1/R2+1/R3)
then Q1=q(1-R1/R2+R2/R3).

b) same thing and the answer is Q1=q(1-R1/R2).

But when he said in question 1 that "S1 is theerafter isolated" and in question 5 "S1 is no more isolated" what is the real difference?(or in other words what is the relationship between s1 and the shell in the two cases?)
In question 4, if we won't add any new conductor to the system,the self capacitances and coefficients of influence C11,C21and C22 stay constant regardless of what changed in the previous parts(like in 3) s2 is earthed but in 2) its not).Is this right?

 

[collinsmark]

OK thanks,I'll show my corrections:
In question 2) b)V∞-V1'=-(_R1∫^R2Edr +_R2∫^R3Edr +_R3∫^∞Edr)
we will get that V1'=Q1/4∏εR1 -Q1/4∏εR2+Q1/4∏εR3

Yes, I think that's correct.

c)V∞-V2=-(_R2∫^R3Edr(0)+_R3∫^∞Edr)
then we will get that V2=Q1/4∏εR3.

Yes, I think that's correct again.

In question 3) a) Electric field outside shell is zero,lets apply gauss law E∫ds=Qin/ε then
E4∏r²=Qinc/ε.(r>R3)
then Qinc=0 and as Qin=Q1+Q2 then Q2=-Q1.(But i think charges on the outer surface of the shell will be zero and -Q2 whice is -Q1 will surround the inner surface the inner surface of the shell).

Yes, very nice.

b)v∞-v1"=-_R1∫^∞Edr
thus -v1"=-(_R1∫^R2Edr +_R2∫^R3Edr (whice is zero since its Ein=0 ) +_R3∫^∞Edr(whice is zero).
we will get that V1"=Q1/4∏ε(1/R1-1/R2).

Yes, that looks right.

5) a) Let q be the charge of s1, the charge distribution on the conductors will be same as previous parts.
V∞-V1=_R1 ∫^R2Edr+_R2∫^R3Edr(0)+_R3∫^∞Edr
then V1=q/4∏εR1 -q/4∏εR2 +q/4∏εR3
Q1/4∏εR1=q/4∏ε(1/R1-1/R2+1/R3)
then Q1=q(1-R1/R2+R2/R3).

Yes. But now your ready to submit the final answer, change q to Q'₁, as indicated in the problem statement (they're the same thing here).

b) same thing and the answer is Q1=q(1-R1/R2).

That looks right. Similarly though, change q to Q''₁ before submitting your final answer.

But when he said in question 1 that "S1 is theerafter isolated" and in question 5 "S1 is no more isolated" what is the real difference?(or in other words what is the relationship between s1 and the shell in the two cases?)

I think what the problem statement means when it says "...and thereafter isolated," really means "...and thereafter isolated, until later, when I tell you something new or different has happened."

By the way, when a conductor is not isolated, it means that net charge can flow into and out of it. It does not necessarily mean that net charge will flow into or out of it, it just means that it can. When a conductor is isolated, any net charge it has (if any) is trapped there, and stays.

In question 4, if we won't add any new conductor to the system,the self capacitances and coefficients of influence C11,C21and C22 stay constant regardless of what changed in the previous parts(like in 3) s2 is earthed but in 2) its not).Is this right?

Yes, I believe so. Capacitance should not change merely by changing charge. So you're right.

On the other hand, in order to calculate the self capacitances, you may wish to add charges so you can calculate the potentials and then solve the capacitance in question by invoking C = Q/V. (And you do this differently depending on which, individual self capacitance or relative capacitance you are trying to calculate).

 

[amjad-sh]

Thanks a lot for your effort collinsmark ,everything is clear now!