### Help with a Physics Question, Gaussian Surfaces.

Posted:

**June 9th, 2008, 4:47 am**Hello guys. This is my first time here and your help would be greatly appreciated!

We have two thin concentric conducting spheres r1 and r2. THe spheres are connected with a power source which supplies emf of epsilon_0. The smaller sphere is connected to the positive pole of the power supply and the larger sphere is connected to the negative pole. Q is given. Using Gauss Law find the Electric Field between r1< R <r2 and find the electric field R>r2.

For this the electric field R. r2 would be 0 correct? Because charge enclosed is 0?

And for the electric field between r1 and r2 it would be equal to E = (+Q)/(4pi*R^2*Epsilon_0)?

After disconnecting the battery we ground the outer shell. Now calculate the electric field R>r2 and R<r1.

For R<r1, the electric field is 0 because charge enclosed is 0, correct?

And for R>r2 the electric field is now equal to E = (+Q)/(4pi*R^2*Epsilon_0) because charge enclosed is now +Q +Q -Q = +Q.

Could you guys please verify if my reasoning is correct for these problems. Thank you so much for your help!

We have two thin concentric conducting spheres r1 and r2. THe spheres are connected with a power source which supplies emf of epsilon_0. The smaller sphere is connected to the positive pole of the power supply and the larger sphere is connected to the negative pole. Q is given. Using Gauss Law find the Electric Field between r1< R <r2 and find the electric field R>r2.

For this the electric field R. r2 would be 0 correct? Because charge enclosed is 0?

And for the electric field between r1 and r2 it would be equal to E = (+Q)/(4pi*R^2*Epsilon_0)?

After disconnecting the battery we ground the outer shell. Now calculate the electric field R>r2 and R<r1.

For R<r1, the electric field is 0 because charge enclosed is 0, correct?

And for R>r2 the electric field is now equal to E = (+Q)/(4pi*R^2*Epsilon_0) because charge enclosed is now +Q +Q -Q = +Q.

Could you guys please verify if my reasoning is correct for these problems. Thank you so much for your help!