Parallel plates capacitor

In summary, the problem deals with finding the potential as a function of position for a set up with capacitor parallel plates separated by a distance d, a potential difference of V_0, and a semi-sphere at the center with a radius of r_0. The solution involves using Gauss' Law and the electric field equation to find the potential for two different regions, one for r<|r_0| and one for r>|r_0|. The problem also suggests using the image charge method to find a more accurate solution.
  • #1
alejandrito29
150
0

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
 

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  • #2
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 !
 
  • #3
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
 

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  • #4
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 ?
 

What is a parallel plates capacitor?

A parallel plates capacitor is a device used to store electrical energy. It consists of two parallel metal plates separated by a small distance and is able to store charge on its plates when connected to a power source.

How does a parallel plates capacitor work?

A parallel plates capacitor works by creating an electric field between its two plates. When a voltage is applied to the capacitor, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field that can store energy in the form of electrical charge.

What factors affect the capacitance of a parallel plates capacitor?

The capacitance of a parallel plates capacitor is affected by the distance between the plates, the surface area of the plates, and the material used for the plates. A larger distance between the plates will result in a lower capacitance, while a larger surface area and a material with a higher dielectric constant will result in a higher capacitance.

What is the formula for calculating the capacitance of a parallel plates capacitor?

The formula for calculating the capacitance of a parallel plates capacitor is C = ε0A/d, where C is the capacitance, ε0 is the permittivity of free space, A is the surface area of the plates, and d is the distance between the plates. This formula assumes that there is a vacuum or air between the plates.

What are some common applications of parallel plates capacitors?

Parallel plates capacitors are used in various electronic devices, such as radios, televisions, and computers. They are also used in power supplies, filters, and timing circuits. Additionally, parallel plates capacitors are used in energy storage systems, such as electric cars and renewable energy sources, to store large amounts of electrical energy.

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