Finding Expressions for Electric Fields with Radius

In summary, the problem involves a solid nonconducting sphere with a uniformly distributed charge of +Q and a conducting layer of gold with a total charge of -2Q on its surface. Using Gauss's Law, the electric field E(r) for r<a (inside the sphere, up to and excluding the gold layer) is expressed as E=Q/4∏r^2ε0. For r>a (outside the coated sphere, beyond the sphere and the gold layer), the electric field is the vector sum of the two charges, but the solution using Gauss's Law is unclear.
  • #1
kgal
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Homework Statement


A solid nonconducting sphere of radius a has a has a total charge +Q uniformly distributed throughout its volume. The surface of the sphere is coated witha avery thing (negligable thickness) conducting layer of gold. A total charge of -2Q is placed on this conducting layer.
Use Gauss's Law to do the following:
a. Find the expression for the electric field E(r) for r<a (inside the sphere, up to and excluding the gold layer).
b. Find the expression for the electric field E(r) for r>a (outside the coated sphere, beyond the sphere and the gold layer).


Homework Equations



Eflux = ∫E dot dA = Q/ε0
A sphere = 4∏r^2

The Attempt at a Solution


I got part a. like this:

Eflux = ∫E dot dA (electric field is constant because uniformly distributed) = EA = E*4∏r^2 = Q/ε0.
solving for E = Q/ 4∏r^2ε0

is this right?

for part b. I am lost...
 
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  • #2
I drew the diagram and it looks like there is two charges (Q and -2Q) so I thought the electric field would be the vector sum of the two. But I am not sure how to solve this using Gauss's Law.Any help is much appreciated!
 

What is an electric field?

An electric field is a physical quantity that describes the ability of an electric charge to exert a force on other charges within its vicinity. It is represented by a vector quantity, with both direction and magnitude.

How is the electric field strength calculated?

The electric field strength is calculated by dividing the force exerted on a charge by the magnitude of the charge itself. This can be represented by the formula E = F/q, where E is the electric field strength, F is the force, and q is the charge.

What is the relationship between electric fields and radius?

The electric field strength is inversely proportional to the square of the distance from the source charge. This means that as the radius increases, the electric field strength decreases.

What is the significance of finding expressions for electric fields with radius?

Finding expressions for electric fields with radius allows us to understand and predict the behavior of electric fields at different distances from a source charge. This is crucial in applications such as electrical engineering and physics.

How can electric fields with radius be used in practical applications?

Electric fields with radius can be used in a variety of practical applications, such as designing and analyzing electronic circuits, calculating the force and motion of charged particles, and understanding the behavior of electromagnetic waves.

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