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Potential inside a hemispherical shell

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i have a hemispherical shell with a radius of R ,and which has V=V0 potential on it. the main problem is i have to find the potential inside the shell for every (r,[tex]\theta[/tex],[tex]\phi[/tex]) points.

1. I think the potential will depent on only the distance from origin, there won't be any [tex]\theta[/tex] or [tex]\phi[/tex] in the equation. on the other hand, if has a charge on it not a potential, inside the shell potential would be constant. So, is it gonna be constant or not in this situation?



2. my attemp to find the potential inside shell is:

i will find the potential on P point (shown in the picture)

V(P)=k[tex]\int[/tex] [tex]\sigma[/tex] dS / r

r=(s2+R2-2Rscos[tex]\theta[/tex])1/2

dS=R2sin[tex]\theta[/tex] d[tex]\theta[/tex]d[tex]\phi[/tex]

for a hemispherical shell [tex]\theta[/tex]=0-pi (?) and [tex]\phi[/tex] = 0-2pi

if i put them in the integral and solve it the result is V(p)=2 [tex]\pi[/tex] kR [tex]\sigma[/tex] (s-R)/(s)

if this is right, i can use V0 and find the result in terms of V0. now if anyone can tell me if this is right or wrong i can continue, i will simulate this event and i really appreciate if anyone help me.

you dont have to solve the integrals, i just need to know if the way of my thinking is right or wrong. if it is wrong which path shoul i follow?
 

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  • #2
gabbagabbahey
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i have a hemispherical shell with a radius of R ,and which has V=V0 potential on it. the main problem is i have to find the potential inside the shell for every (r,[tex]\theta[/tex],[tex]\phi[/tex]) points.

1. I think the potential will depent on only the distance from origin, there won't be any [tex]\theta[/tex] or [tex]\phi[/tex] in the equation. on the other hand, if has a charge on it not a potential, inside the shell potential would be constant. So, is it gonna be constant or not in this situation?



2. my attemp to find the potential inside shell is:

i will find the potential on P point (shown in the picture)

V(P)=k[tex]\int[/tex] [tex]\sigma[/tex] dS / r

r=(s2+R2-2Rscos[tex]\theta[/tex])1/2

dS=R2sin[tex]\theta[/tex] d[tex]\theta[/tex]d[tex]\phi[/tex]

for a hemispherical shell [tex]\theta[/tex]=0-pi (?) and [tex]\phi[/tex] = 0-2pi

if i put them in the integral and solve it the result is V(p)=2 [tex]\pi[/tex] kR [tex]\sigma[/tex] (s-R)/(s)

if this is right, i can use V0 and find the result in terms of V0. now if anyone can tell me if this is right or wrong i can continue, i will simulate this event and i really appreciate if anyone help me.

you dont have to solve the integrals, i just need to know if the way of my thinking is right or wrong. if it is wrong which path shoul i follow?
In problems where you aren't given the charge density, but are given the potential along the boundary of a region, and asked to find the potenial everywhere inside the region, you want to solve Laplace's equation (assuming there is no charge inside the region). In this case, the easiest method to use is Separation of Variables in Spherical coordinates. The general solution is well known and should be in your text. You need only apply your boundary condition(s) and you are done.
 
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ok then i will try to use separation of variables,
i know the general solution for sphere shell. it should be like (Al . r^l + Bl. r ^(l+1) ) . Pl(cos) , but can I use this for hemisphere or should i find a special solution for hemisphere.
 
  • #4
gabbagabbahey
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ok then i will try to use separation of variables,
i know the general solution for sphere shell. it should be like (Al . r^l + Bl. r ^(l+1) ) . Pl(cos) , but can I use this for hemisphere or should i find a special solution for hemisphere.
You should be able to use it for your hemisphere just fine. You need only specify your boundary condition(s) properly in Spherical Coordinates (Careful, if your hemisphere has a bottom disk at potential [itex]V_0[/itex] that has to be part of your boundary condition).
 
  • #5
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ok thank you very much , i will try to solve it with bndry cond. now
 
  • #6
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You should be able to use it for your hemisphere just fine. You need only specify your boundary condition(s) properly in Spherical Coordinates (Careful, if your hemisphere has a bottom disk at potential [itex]V_0[/itex] that has to be part of your boundary condition).
i solved it, and no there is no bottom of the hemisphere. i used
V(r,[tex]\theta[/tex])= Al rl Pl (cos[tex]\theta[/tex])

and since i have V(R,[tex]\theta[/tex]) = V0 only term i have is A0

and i find that A0=V0. so Vin=V0 i found.
do you think it make sense to have the same potential everywhere ?
 
  • #7
gabbagabbahey
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do you think it make sense to have the same potential everywhere ?
No, if there is no bottom to the hemispherical shell (it is a thin shell right?), then the surface isn't closed and your other boundary is at infinity. As you approach infinity, you would expect the potential to go to zero, not stay at [itex]V_0[/itex].

In any case, you've gotten your other boundary condition wrong as well. [itex]V(R,\theta)=V_0[/itex] only for [itex]0\leq \theta \leq \pi/2[/itex]. What you've actually calculated is the potential inside an entire spherical shell held at [itex]V_0[/itex].
 
  • #8
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No, if there is no bottom to the hemispherical shell (it is a thin shell right?), then the surface isn't closed and your other boundary is at infinity. As you approach infinity, you would expect the potential to go to zero, not stay at [itex]V_0[/itex].

In any case, you've gotten your other boundary condition wrong as well. [itex]V(R,\theta)=V_0[/itex] only for [itex]0\leq \theta \leq \pi/2[/itex]. What you've actually calculated is the potential inside an entire spherical shell held at [itex]V_0[/itex].
yes it is thin and surface isnt closed. and yes i did take the boundry cond. wrong.

so ? i have made some calculation and i am not sure i did it right, you see i am not good at emt calculations. i have found potential and then electric field inside shell for a point, then run some calculations with matlab, and electric field didnt make any sense inside the shell, so i thought i did something wrong, obviously. can you tell me my mistake, again? thank you.
 

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