Electric field outside two concentric sphere's

[bulbanos]

We've got two concentric metallic spheres, let's say P and Q
P is the smallest one, so its inside Q
Q is grounded and P is positive charged

I figured out that the field in between P and Q is not homogeneous but constant. But why is the field outside Q zero while it is not inside Q? Still the field is zero inside P because of the Gauss' Law...
I just don't see the profound difference between the two spheres.

 

[vincentchan]

the field in between P and Q is not homogeneous but constant.

You sure it is constant? I don't think so... a paralleled metal plate will riase a constant E field in between, but not for a concentric sphere.. You may want to check your calcultion, the field should be similar to a point charge in free space

But why is the field outside Q zero while it is not inside Q? Still the field is zero inside P because of the Gauss' Law...

The field inside the Q is not zero is also because of gauss law.. draw an gaussian surface in between P and Q, the charge enclosed in the gaussian surface is surely not zero, [tex] Q_{enclosed}/ \epsilon = \int \vec{E} d \vec{S} [/tex].. so how could [tex] \vec{E} [/tex] be zero?

why is the field outside Q zero

Q carries negative charge and P carries positive charge.. and there E field cancels out.. Question: Where did the negative charge come from?

 

[clive]

If "constant" means "does not vary in time" (after equilibrium, obviously) then bulbanos is right. After equilibrium there is an electroSTATIC field.

The external sphere is grounded then V_Q=0. But this sphere is in the electrical field created by P (q/4/pi/eps_0/R_Q). R_Q is the radius of external sphere In order to have a zero potential, external sphere will get -q from Earth and will have its own potential -q/4/pi/eps_0/R_Q. The total potential of Q will be now given by the sum between the potential of the field from P and the potential from its own field that is 0.

In any external point, the field is zero because the electrical charge inside the Gaussian surface containing this point is 0: q+(-q)=0. (see the post of vincentchan for the Gauss' law)