[E&M] Question on the Image charge method of a grounding sphere

[Shell_E]

So I was reading Jackson's discussion on Image charge method of a grounding sphere.

He first assumed an image charge q inside Sphere with radius a, so the potential for real change and image charge is

.
The by set potential equal to 0 at x=a, he solved q' and y'

Then he can get potential, thus get E field, also can get force on real charge q.
Then he figured out surface charge density by:

Then he mentioned that the force on unit area da is just :

So this is the part I don't get, I understand that σ *da = dq is the charge on area da, but what is the rest part? I don't really get why dF = dq * (σ /2ε ), is that indicate the E field at da is (σ /2ε )? if so, how do we get this?

Thank you!

 

[mitochan]

Hi.

1. Say on thin volume surface surrounding da on conductor sphere,
$E=\frac{\sigma}{\epsilon_0}$ outside surface and $E=0$ inside surface.

2. Maxwell stress tensor says force per unit area on outside surface is $\frac{\epsilon_0}{2} E^2=\frac{\sigma^2}{2\epsilon_0}$

3. 1, 2 and $dF=\sigma da E$ tempt us to regard

E field at da is (σ /2ε )?

value of which is exact average of outside field and inside field mentioned in 1.

I do not think electric field at the exact point of sphere charge be defined neither physically nor mathematically.

 

[Delta2]

if you integrate coulomb's law (i.e. the sum of all coulomb forces from all the other elements da of the sphere to a particular element da') to calculate the total force on a surface element da' of the sphere that's the result that you will get. You can see this at this thread
https://www.physicsforums.com/threa...-spherical-shell-is-zero.977305/#post-6233257

 

[mitochan]

Hi.
Further to post #2 I found texts around Fig.I.5 in Introduction of Jackson Third edition is helpful showing microscopic and macroscopic view of plate charge. It says about plate condenser but applies to our sphere.

Though I said electric field working on sphere charge is not defined, we may make use of convention that
$$E=H_{1/2}(x) \frac{\sigma}{\epsilon_0}$$
where Heaviside step function_1/2 is 0 for x<0 inside, 1/2 for x=0 on surface and 1 for x>0 outside. As Jackson explained it does not correspond to both macro and micro physics and I take it for practical mathematical convention to meet equation $dF=\frac{E}{\epsilon_0} \sigma da$.