# Electric Field in Concentric Spheres

1. Feb 7, 2010

### 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!

Last edited: Feb 7, 2010
2. Feb 7, 2010

### xcvxcvvc

To use that equation for a sphere of charge, you must understand where the equation came from. It is from Gauss's law.
$$\int_A E dA=\frac{Q_{enc}}{\epsilon_o}$$
$$E\int_A dA=\frac{Q_{enc}}{\epsilon_o}$$

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:
$$E(4\pi r^2)=\frac{Q_{enc}}{\epsilon_o}$$
$$E=\frac{Q_{enc}}{4\pi\epsilon_or^2}$$

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.

Last edited: Feb 7, 2010
3. Feb 7, 2010

### 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?

4. Feb 7, 2010

### 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.

5. Feb 8, 2010

### aifan27

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