How Does a Spherical Inclusion Affect Potential in a Parallel Plate Capacitor?

[alejandrito29]

Homework Statement

Capacitor parallel plates separation $$d$$, potencial diference $$V_0$$, at the center there is a semi sphere of radius $r_0$.

Find the potencial as function of position if $$d>>r_0$$

Homework Equations

I think that relevant equation are Gauss Law

The Attempt at a Solution

Mi proposed of solution is

$$V_{plates}= \sigma /\epsilon_0 z $$

for $$r<|r_0|$$

Gauss Low

$$E 4/3 \pi r^2 = \sigma 4/3 \pi r_0^2$$

$$E= \sigma r_0^2 /\epsilon_0 r^{-2} \hat{r} $$

$$ V_{sp} = \sigma r_0 /\epsilon_0 ( r_0/r-1) = \sigma r_0 /\epsilon_0 ( r_0/\sqrt{x^2+y^2+z^2}-1) $$

And for $$r>|r_0|$$

$$ V=V_{plates}+V_{sp}$$

Help please

 

[BvU]

Hi Alex,

Can't really follow. It looks to me as if you assume $\sigma$ is a constant, but why would it be constant ? The plates are conducting, so the electric field has to be perpendicular to the surface at all places. I don't think you potential satisfies that.

Note that your addition counts in a circular disk radius r0 that isn't really there !

 

[alejandrito29]

Tanks

Some suggest for solving my problem?

my adittion out of radius $$r_0$$ is because there is two field electrics at zone $$r>|r_0|$$, then two potential . I don't am sure of this.

Help please

 

[BvU]

I think I understand what you mean, and I tried to bring across that the complete field is not a simple addition obtained from two plates and half a sphere. You will have to find something that satisfies the field equations and the boundary conditions as well. Not so trivial, I grant you. Do you know about the image charge method ?