Finding Electric Field due to hemispherical charge distribution

  • #1
S[e^x]=f(u)^n
23
0

Homework Statement


I need to find the electric field along the Z-axis for a hemispherical open shell of radius R (open face down). the shell is oriented so its symmetrical about the Z axis.


The Attempt at a Solution


What i've done is set it up, taken the vector from the origin (at the center of what would be the full sphere), and subtracted the vector bound to the z axis, to get the vector from the differential surface of the shell to an arbitrary point on the z axis. I then integrate, using spherical coordinates the 1/r^2 expression...

I can't seem to get this to work, can someone guide me through the right steps?, i've been trying to do this for about 7hrs now, and feel really stupid since i'm a 3rd year uni student majoring in physics....
 

Answers and Replies

  • #2
cupid.callin
1,137
0
consider any thin circular strip on hemisphere at angle θ from vertical and subtending angle dθ on center.

find field due to this just like you do for ring on axis
then integrate it from θ = 0 to 90 degree
 
  • #3
S[e^x]=f(u)^n
23
0
I've tried that, and i end up with E= [(surface charge density)*(R)^2]/[2*epsilon*(R-z)^2]

where R is the hemispheres radius, and z, is a point on the z axis...

i then integrate this to find the potential between R (along the z axis), and the origin, which ends up in a divide by 0 error... so i know the Electric field equation cannot be right although it seems to check out dimensionally...

is there an easier way to find the potential between the origin and R(on the z axis) that i'm missing? right now i'm just going after the electric field, which should then lead to a simple integration to get the potential difference between the two points. or have i forgotten something?
 
  • #4
gneill
Mentor
20,945
2,886
S[e^x]=f(u)^n;3119613 said:
I've tried that, and i end up with E= [(surface charge density)*(R)^2]/[2*epsilon*(R-z)^2]

where R is the hemispheres radius, and z, is a point on the z axis...

Perhaps you could show how you arrived at that? The well-known result for the on-axis field produced by a ring of charge Q of radius a and distance x is given by:

[tex]E_x = \frac{{kQx}}{{\left( {x^2 + a^2 } \right)^{{3 \mathord{\left/
{\vphantom {3 2}} \right.
\kern-\nulldelimiterspace} 2}} }}[/tex]
 

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