Magnitude of the Electric Field at the Following Distances

In summary, the problem requires finding the magnitude of the electric field at various radial distances from a spherical shell with uniform volume charge density ρ = 1.87 nC/m3, inner radius a = 15.0 cm, and outer radius b = 2.60a. Two approaches are suggested: using the shell theorem or considering the superposition of the electric fields from two concentric solid spheres. The latter approach involves finding the field due to a solid sphere of charge density -ρ and radius a, and the field due to a solid sphere of charge density +ρ and radius b, and adding them together. Reference material for solving electric fields due to spherical shells is also provided.
  • #1
Darkgora

Homework Statement


The figure below shows a spherical shell with uniform volume charge density ρ = 1.87 nC/m3, inner radius a = 15.0 cm, and outer radius b = 2.60a.

[Reference Picture]

What is the magnitude of the electric field at the following radial distances?

Homework Equations


E=k(q/r^2)
E=k(q/R^3)*r^2

The Attempt at a Solution


I only understand how to compute an electric field for a solid sphere. I am unsure how to alter the given equation to compensate for the cavity within the charged spherical shell.
phy 1.png
 
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  • #2
The shell theorem is used to answer questions like this. It enables one to ignore spherically-distributed charge at distance greater than r from the centre of the sphere, and the rest of the charge can be assumed to be at the centre of the sphere..
 
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  • #3
Here is an alternative approach. Since you understand the electric field due to a solid sphere, for this problem consider the superposition of the electric fields from two concentric solid spheres, one of radius b and charge density +ρ and one of radius a and charge density -ρ.
 
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  • #4
kuruman said:
Here is an alternative approach. Since you understand the electric field due to a solid sphere, for this problem consider the superposition of the electric fields from two concentric solid spheres, one of radius b and charge density +ρ and one of radius a and charge density -ρ.

I am unsure how to compute the electric field given charge density, thus I have solved for the total charge enclosed:

Volume of hollow sphere * Charge Density
= 4(pi)/3(.351^3-.13^3) * (1.89e-9)

When I plug this total enclosed charge value into kqr/R^3 and subtract the inner from outer sphere I am not getting the right answer.
 
  • #5
Darkgora said:
When I plug this total enclosed charge value into kqr/R^3 and subtract the inner from outer sphere I am not getting the right answer.
You missed the point of my suggestion. Follow these steps for the field in parts d -f.
1. Find the field due to a solid sphere of charge density -ρ and radius a at that radius.
2. Find the field due to a solid sphere of charge density +ρ and radius b at that radius.
3. Add the fields (one is positive the other negative).

Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html
 

1. What is the formula for calculating the magnitude of the electric field at a given distance?

The formula for calculating the magnitude of the electric field at a given distance is E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the source, and r is the distance between the source and the point of interest.

2. How does the magnitude of the electric field change with distance?

The magnitude of the electric field is inversely proportional to the square of the distance. This means that as the distance increases, the magnitude of the electric field decreases.

3. What is the unit of measurement for the magnitude of the electric field?

The unit of measurement for the magnitude of the electric field is Newtons per Coulomb (N/C).

4. How does the magnitude of the electric field vary for different sources?

The magnitude of the electric field varies depending on the charge of the source. The higher the charge, the stronger the electric field will be at a given distance.

5. How can the magnitude of the electric field be used to determine the direction of the field?

The magnitude of the electric field alone cannot determine the direction. To determine the direction, one must also know the direction of the source's electric field (positive to negative) and the direction of the test charge in the electric field (negative to positive).

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