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  • (a) Apply Gauss’s law to show that for a charged spherical shell, the electric field outside the shell is, as if the entire charge were concentrated at the center. (b) What would be the elect

(a) Apply Gauss’s law to show that for a charged spherical shell, the electric field outside the shell is, as if the entire charge were concentrated at the center. (b) What would be the electric flux through a sphere which contains an electric dipole?

 

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How to Use Gauss's Law on a Charged Spherical Shell (a)

This is what Gauss's law says:

?E⋅dA = Qenc/?0

where:

The electric field is denoted by E.

The very small surface area part is denoted by dA.

The charge inside is called Qenc.

The permittivity of empty space is ?0.

First, look at a Gaussian surface.
We use a circular Gaussian surface with a radius r (which is bigger than the shell's radius) and center it on the same point as the charged shell.

Step 2: Think about symmetry
So, the electric field must be uniform and spread out in a general direction across the whole Gaussian surface. This means that we can take E out of the integral:

E(4πr2)=E(4πdA)

Step 3: Use Gauss's Law.
This Gaussian surface holds a total charge of Q, which is the total charge of the shell. Using the law of Gauss:

E(4πr²) = Q/??

This is the same thing as the electric field of a point charge Q in the middle of the shell. So, the field acts outside the shell as if all the charge were gathered in the middle.


The flow of electricity through a sphere that has an electric dipole

Gauss's law tells us how much electricity flows through a closed surface:


ΥE=?E⋅dA=Qenc/?0

If you have an electric dipole with a positive charge (+q) and a negative charge (−q), the total charge inside any closed surface around the dipole is

Qenc=(+q)+(-q)

So, the net flow of electricity through a surface that is closed around an electric dipole is

E = 0

In other words, even though an electric dipole creates an electric field, there is no net flow because the same number of field lines are coming into and leaving the surface.


 

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