Two conducting hollow spheres, a point charge and electric field prob

AI Thread Summary
A point charge is located at the center of a conducting hollow sphere, which is within another hollow sphere, creating a complex electric field scenario. The inner sphere has a negative charge while the outer sphere is positively charged, affecting the electric field at point A, located between them. Gauss' Law is crucial for determining the electric field, as it states that the total flux through a closed surface is equal to the net charge enclosed. The electric field at point A results from the contributions of the point charge and the inner sphere, while the outer sphere does not contribute to the field inside it. The net electric field can be calculated by considering the positive contribution from the point charge and the negative contribution from the inner sphere.
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Homework Statement


A point charge is at the center of a conducting hollow sphere, with radius 0.011m, that is within another conducting hollow sphere of radius 0.041m. The point charge is Q0=+4.30e-6 C, the inner sphere has a net charge of Q1=-1.70e-6 C, and the outer sphere has a net charge of Q2=+6.50e-6 C.
Calculate the magnitude of the electric field at a point A located 0.021m from the center.

Homework Equations


Gauss' Law \Phi= Qin/ \epsilon = E*Areas
Areas=4\pir2
\epsilon=permittivity constant

The Attempt at a Solution


I've been stuck on this for awhile and don't really know where to begin. I know point A is between the two hollow spheres, one of which is positive, the other negative. But I don't know what to think of the conducting spheres when they aren't in electrostatic equil because they don't have net Q=0. Will the E of point A just be the sum of the electric fields of the spheres and point charge?
EA= E0+E1+E2?
 
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Think about the implications of Gauss Law.

The total flux through a closed surface is going to be the net charge within.

So at P the field you are looking at will be from the point charge and the inner sphere. The outer sphere since you are inside it ... what does that contribute?
 
It won't contribute anything.

So I tried KQ0/(distance to A)2 - KQ1/(distance to A)2 and it wasn't right.

What's up?
 
yes it would. you have a positive flux due the point charge and a negative flux due to the inner center. if you can find the net flux, you can find the net electric field. (also, think of A as being an invisible circle instead of a point.)
 
shelbz23 said:
yes it would. you have a positive flux due the point charge and a negative flux due to the inner center. if you can find the net flux, you can find the net electric field. (also, think of A as being an invisible circle instead of a point.)

alright i got it all figured out. s the net charge only the magnitudes or would you subtract the negative charge?
 
You would get a positive contribution from Q0 and a negative contribution from Q1.
 
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