Electric field

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JacksonCzarniak09
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Electric field

Post by JacksonCzarniak09 »

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
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Rich Feldman
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Re: Electric field

Post by Rich Feldman »

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!
All models are wrong; some models are useful. -- George Box
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Liam David
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Re: Electric field

Post by Liam David »

The calculation is very quick if you know Coulomb's law and how the electric field and electric potential are related.
PXL_20240519_032447676.jpg
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://www-pnp.physics.ox.ac.uk/~gwenla ... hs-4ed.pdf). The material in chapters 1-2 will give you the tools to analyze situations like this.
JacksonCzarniak09
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Re: Electric field

Post by JacksonCzarniak09 »

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?
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Rich Feldman
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Re: Electric field

Post by Rich Feldman »

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.phy-astr.gsu.edu/hb ... otsph.html
Image
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
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Rich Feldman
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Re: Electric field

Post by Rich Feldman »

Sorry, Liam, I forgot to credit your back-of-envelope 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 10-cm 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
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