# Non-conductve sphere with cavity -- find Electric field

• Manolisjam
In summary: Keep the vollume inside the integral!In summary, the enclosed charge will be the sum of the charges in those shells "below" the radius of interest.
Manolisjam
I have a non conducting sphere with a charge ρ=A/r per uni vollume A is constant. suppose there is a cavity in the centre and within a particle of charge q. i want to find the E inside the sphere in respect with r.

## The Attempt at a Solution

for radius equal of the cavity i get ##E=kq/r^2## for r bigger than the radius of the sphere ##E=k(Q+q)/r^2## now only one case left
if r is bigger than the radius of the cavity and smaller than the radius of the sphere.
##E4πr^2=\frac{q+ρV}{ε_0} ##
where ρ is the density per unit vollume and V is the vollume
is my enclosed Q right? and my V is 4π(r^3-a^3) where a^3 is the small radius.

Last edited by a moderator:
Manolisjam said:
is my enclosed Q right?
Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.

gneill said:
Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.
?

Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.

gneill said:
Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.
##E4πr^2=\frac{q+ \int ρVdr}{ε_0} ## is the right one?

Manolisjam said:
##E4πr^2=\frac{q+ \int ρVdr}{ε_0} ## is the right one?
That's the idea, yes.

gneill said:
That's the idea, yes.
so i Keep the vollume inside the integral!

You want to integrate over the volume of interest, summing up the charges contained in each differential volume element. So really that V should be a dV(r). I assumed (my bad) that that was what you were implying with that V.

## What is a non-conductive sphere with a cavity?

A non-conductive sphere with a cavity is a spherical object made of a material that does not allow electricity to flow through it, and it has a hollow space inside.

## Why is it important to find the electric field in a non-conductive sphere with a cavity?

Finding the electric field in a non-conductive sphere with a cavity is important because it helps us understand how charges are distributed within the cavity and how they interact with the surrounding material.

## How do you calculate the electric field in a non-conductive sphere with a cavity?

The electric field in a non-conductive sphere with a cavity can be calculated using the Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space.

## What factors affect the electric field in a non-conductive sphere with a cavity?

The electric field in a non-conductive sphere with a cavity is affected by the charge enclosed by the cavity, the size and shape of the cavity, and the permittivity of the surrounding material.

## How does the electric field inside the cavity of a non-conductive sphere differ from the electric field outside the sphere?

The electric field inside the cavity of a non-conductive sphere is not affected by the charges on the surface of the sphere, while the electric field outside the sphere is affected by the charges on the surface. Additionally, the electric field inside the cavity is constant, while the electric field outside the sphere decreases as the distance from the sphere increases.

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