Show that the electric field outside the shell is everywhere

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The electric field outside a uniformly charged hollow spherical shell is equivalent to the field produced by a point charge Q at the center of the shell, as demonstrated using Gauss's Law. By selecting a spherical Gaussian surface, the enclosed charge is identified as Q, leading to the relation E * 4πr² = Q/ε0. This allows for the derivation of the electric field as E = Q/4πε0r², confirming its uniformity in magnitude and direction. Consequently, the electric field outside the shell remains consistent at all points. Thus, Gauss's Law effectively proves the equivalence of the electric fields in this scenario.
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A hollow spherical shell is uniformly charged with a total charge Q. Show that the electric field outside the shell is everywhere the same as the field due to a point charge Q located at the center of the shell.
 
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As in your other thread, just use Gauss's Law. Do you need help applying Gauss's Law?

Just for a sidenote, be happy that you get to use Gauss's Law for this. Proving it without it is... ugly, to say the least.

cookiemonster
 


The electric field outside the shell can be shown to be everywhere the same as the field due to a point charge Q located at the center of the shell by using Gauss's Law.

Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). In this case, we can choose a spherical Gaussian surface with radius r, centered at the center of the shell.

Since the shell is uniformly charged, the electric field at any point on the surface of the shell must be perpendicular to the surface. This means that the electric field is constant in magnitude and direction on the surface of the shell.

Using Gauss's Law, we can write:

∫E⃗⋅dA⃗ = Qenc/ε0

Where E⃗ is the electric field, dA⃗ is the area element of the Gaussian surface, and Qenc is the enclosed charge. Since the electric field is constant on the surface of the shell, we can pull it out of the integral and are left with:

E⃗ ∫dA⃗ = Qenc/ε0

The integral on the left hand side is simply the surface area of the Gaussian surface, which is 4πr^2. The enclosed charge, Qenc, is equal to the total charge Q of the shell.

Therefore, we can write:

E⃗ 4πr^2 = Q/ε0

Solving for the electric field, we get:

E⃗ = Q/4πε0r^2

This is the same expression for the electric field due to a point charge Q located at the center of the shell. This proves that the electric field outside the shell is everywhere the same as the field due to a point charge Q located at the center of the shell.

In conclusion, using Gauss's Law, we have shown that the electric field outside the shell is everywhere the same as the field due to a point charge Q located at the center of the shell. This means that the electric field outside the shell is constant in magnitude and direction at all points, making it uniform.
 
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