Hello,
I am wondering how I can calculate the electric field on the surface of a sphere? I want to do this so I can find the radius needed to have a electric field of around 1 x 10/6 v/m at 50 kV. Thanks
Electric field

 Posts: 21
 Joined: Mon Jan 15, 2024 8:56 pm
 Real name: Jackson Czarniak
 Rich Feldman
 Posts: 1482
 Joined: Mon Dec 21, 2009 6:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Electric field
Chapter 1 in physics textbook about electromagnetism.
Or Internet search (we used to say LMGTFY). I pasted your question, word for word, into a search window. Got answers with more explanation and insight than anyone is likely to contribute here. Now if you ran the formula and got a radius value that surprised you, and want a check on it, that would be reasonable to ask.
On Quora these days your question would be relatively respectable & not invite snarky responses.
Good luck!
Or Internet search (we used to say LMGTFY). I pasted your question, word for word, into a search window. Got answers with more explanation and insight than anyone is likely to contribute here. Now if you ran the formula and got a radius value that surprised you, and want a check on it, that would be reasonable to ask.
On Quora these days your question would be relatively respectable & not invite snarky responses.
Good luck!
All models are wrong; some models are useful.  George Box
 Liam David
 Posts: 548
 Joined: Sat Jan 25, 2014 5:30 pm
 Real name: Liam David
 Location: PPPL
Re: Electric field
The calculation is very quick if you know Coulomb's law and how the electric field and electric potential are related.
Also note that the curvature k = 1/a, so we can write E(a) = k*V_0. Hence the higher the curvature, the higher the surface electric field. This is essentially why sharp corners create strong electric fields, and why you want things that are floating at high voltage to be smooth.
I would encourage you to read an introductory E&M textbook, e.g. Griffiths (http://wwwpnp.physics.ox.ac.uk/~gwenla ... hs4ed.pdf). The material in chapters 12 will give you the tools to analyze situations like this.
Also note that the curvature k = 1/a, so we can write E(a) = k*V_0. Hence the higher the curvature, the higher the surface electric field. This is essentially why sharp corners create strong electric fields, and why you want things that are floating at high voltage to be smooth.
I would encourage you to read an introductory E&M textbook, e.g. Griffiths (http://wwwpnp.physics.ox.ac.uk/~gwenla ... hs4ed.pdf). The material in chapters 12 will give you the tools to analyze situations like this.

 Posts: 21
 Joined: Mon Jan 15, 2024 8:56 pm
 Real name: Jackson Czarniak
Re: Electric field
In the formula i found, E=kQ/r^2, since Q is charge, do i simply find that by multiplying current by time ( 1 second), so in my case 50 mA will be .05 c?
 Rich Feldman
 Posts: 1482
 Joined: Mon Dec 21, 2009 6:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Electric field
Nice desktop there, Liam!
Jackson, I don't know what current (50 mA) or time (1 second) have to do with the problem you stated. (V and E are given; what is R?)
It turns out we don't need to solve for charge Q.
Aforementioned Internet search led right to this Hyperphysics page:
http://hyperphysics.phyastr.gsu.edu/hb ... otsph.html
It gives the potential V=kQ/r and the field strength E=kQ/r^2, with the same k.
So V/E = r.
If I got this right, a radial E field which reaches 1 MV/m at r=0.05 m will have a potential of 50 kV at that radius.
It works so simply because the derivative of 1/r is 1/r^2.
We can follow the lead of the hyperphysics page editor, and omit the rigor of vector and sign notation.
Jackson, I don't know what current (50 mA) or time (1 second) have to do with the problem you stated. (V and E are given; what is R?)
It turns out we don't need to solve for charge Q.
Aforementioned Internet search led right to this Hyperphysics page:
http://hyperphysics.phyastr.gsu.edu/hb ... otsph.html
It gives the potential V=kQ/r and the field strength E=kQ/r^2, with the same k.
So V/E = r.
If I got this right, a radial E field which reaches 1 MV/m at r=0.05 m will have a potential of 50 kV at that radius.
It works so simply because the derivative of 1/r is 1/r^2.
We can follow the lead of the hyperphysics page editor, and omit the rigor of vector and sign notation.
All models are wrong; some models are useful.  George Box
 Rich Feldman
 Posts: 1482
 Joined: Mon Dec 21, 2009 6:59 pm
 Real name: Rich Feldman
 Location: Santa Clara County, CA, USA
Re: Electric field
Sorry, Liam, I forgot to credit your backofenvelope analysis with also showing (right at the top)
formulas for E and V that differ only by the factor r.
And practical formula showing E as product of V and curvature 1/r.
Liam and the hyperphysics page both start with E and derive V by integration. I think that's more fundamental than starting with V and taking its derivative for E.
If we carry things through for fun, the numbers all line up. Q is 278 nC.
The capacitance of 10cm ball is 5.56 pF, matching Q/V ratio and the penultimate formula on Liam's envelope.
Jackson's 50 mA could charge it to 50 kV in 5.56 microseconds.
formulas for E and V that differ only by the factor r.
And practical formula showing E as product of V and curvature 1/r.
Liam and the hyperphysics page both start with E and derive V by integration. I think that's more fundamental than starting with V and taking its derivative for E.
If we carry things through for fun, the numbers all line up. Q is 278 nC.
The capacitance of 10cm ball is 5.56 pF, matching Q/V ratio and the penultimate formula on Liam's envelope.
Jackson's 50 mA could charge it to 50 kV in 5.56 microseconds.
All models are wrong; some models are useful.  George Box