Electric Field in Concentric Spheres

aifan27

1. The problem statement, all variables and given/known data
Two charges concentric spheres have radii of 10.0 cm and 15.0 cm. The charge on the inner sphere is 4.00 x 10 ^-8 C, and that on the outer sphere is 2.00 x 10^-8 C. Find the electric field (a) at r = 12.0 cm and (b) at r = 20.0 cm.


2. Relevant equations
I know that this is a fairly simple problem, but am somewhat confused on which numbers I use. Using this equation:



what should I be using for "Q"? At r = 12.0 cm do I use 4.00 x 10^-8 and at r = 20.0 cm do I use 2.00 x 10^-8? Or am I adding the two charges together to get a total charge, since both charges affect each point.

Thanks!

xcvxcvvc

To use that equation for a sphere of charge, you must understand where the equation came from. It is from Gauss's law.
[tex]\int_A E dA=\frac{Q_{enc}}{\epsilon_o}[/tex]
Your E field is constant at all points on your surface, so your integral becomes:
[tex]E\int_A dA=\frac{Q_{enc}}{\epsilon_o}[/tex]

That integral, I hope you see, becomes the total area of your Gauss surface. What is your surface? It is a sphere, so it becomes the surface area of a sphere:
[tex]E(4\pi r^2)=\frac{Q_{enc}}{\epsilon_o}[/tex]
[tex]E=\frac{Q_{enc}}{4\pi\epsilon_or^2}[/tex]

Where Q is the enclosed charge. Does this answer your question?

If you do not know what a Gauss surface is, just think of it this way roughly: it's an imaginary surface of any shape(yet often tactically chosen for mathematical ease) that you wrap around a charge to calculate the E field produced by that charge along the surface. So in this problem, we are wrapping an imaginary sphere at r = 12 around the charge. Only the enclosed charge influences the E field. Next, we make an imaginary sphere at r =20. Again, only the enclosed charge adds to the field(and a positive charges cancel negative charges too).

Sorry if this explanation is wordy and not to the point. i'm tired, unable to think well, and am going to bed.

aifan27

Not really - at both points (12 cm and 20 cm) away, the charge from both spheres affect it, so I should total the charges and therefore use for Qenc, correct?

xcvxcvvc

for r = 12cm you only use the charge from the inner sphere. At r = 20cm you use both charges added. Q enc stands for enclosed. You can visualize that a sphere of radius 12 cm only encloses the charge from the inner sphere(meaning the outer sphere does not affect E) and that another imaginary sphere at r = 20 encloses both charges, so you sum them since they both affects the E field.


Wow sorry for not noting this earlier. This is how you should think about it: the E field in a conductor is zero. Therefore, only the inner sphere affects the E in the area between the two spheres. However, once you exit the outer sphere, both charges affects the E field.

aifan27

Ok, and "r" is the distance the point is away from the center, not the radius of the circle, right?