Electric Field in a tiny cubical cavity within three piled disks of charge.

In summary, we are given a charge distribution in the shape of a large disk with three layers of uniform densities and a cubical cavity within the layer with density \rho_{2}. The problem asks us to find the electrostatic field at the center of the cubical cavity. Two approaches are suggested, one involving direct calculation of potential and the other using the superposition principle. The issue with both approaches is determining the potential due to the layer with the cavity and calculating the normal derivative of potential. The suggestion is to consider the effect of superimposing a cube of charge density -\rho_2 onto the solid 3 layered disk.
  • #1
ELESSAR TELKONT
44
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Homework Statement



A charge distribution has the shape of a very large disk of thickness [tex]3d[/tex]. This disk has three layers of uniform densities [tex]\rho_{1},\rho_{2},\rho_{3}[/tex] and thickness [tex]d[/tex] each one. Within the layer with density [tex]\rho_{2}[/tex] exists a tiny cubical cavity in such a manner that two of its faces are parallel to the interfaces between layers. Find the electrostatic field at the center of the cubical cavity.

Homework Equations





The Attempt at a Solution



I have tried to solve it basically in two manners. Directly calculating the potential due to the disk layers and using the fact that
[tex]\varphi(x)=\frac{1}{4\pi\epsilon_{0}}\int_{V}\frac{\rho}{\left\vert r-r'\right\vert}dV'+\frac{1}{4\pi}\int_{S}\left [ -\phi \frac{r-r'}{\left\vert r-r'\right\vert^{3}}+\frac{\nabla '\phi}{\left\vert r-r'\right\vert}\right ]\cdot n' dS'[/tex]. My problem in both strategies is that how can I calculate the potential due to the layer where the cubical cavity is and in the second form is how to calculate the normal derivative of potential.

Is there any technic to do this problem easier?
 
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  • #2
Why not just use the superposition principle?
 
  • #3
How, give me a hint
 
  • #4
What happens if you superimpose a cube of charge density [itex]-\rho_2[/itex] onto a solid 3 layered disk (like the one you describe in your problem statement, but without the cavity)?
 

FAQ: Electric Field in a tiny cubical cavity within three piled disks of charge.

What is an electric field?

An electric field is a physical quantity that describes the force exerted by electrically charged particles on other charged particles. It is represented by a vector that has both magnitude and direction.

How is the electric field calculated in a tiny cubical cavity within three piled disks of charge?

The electric field in this scenario can be calculated by using the principle of superposition, which states that the total electric field at a point is the vector sum of the individual electric fields produced by each charged object. The formula for electric field is E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the object, and r is the distance from the object.

What factors affect the strength of the electric field in a tiny cubical cavity within three piled disks of charge?

The strength of the electric field is affected by the magnitude and distribution of charge within the three piled disks, as well as the distance from the charged objects. The shape and size of the cubical cavity may also play a role in determining the electric field.

What are the units of electric field?

The SI unit for electric field is Newtons per Coulomb (N/C). However, other commonly used units include Volts per meter (V/m) and Pascals (Pa).

How is the direction of the electric field determined in a tiny cubical cavity within three piled disks of charge?

The direction of the electric field is determined by the direction in which a positive test charge would move if placed at a particular point. The field lines always point away from positively charged objects and towards negatively charged objects.

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