# Net charge on sphere

1. Feb 6, 2006

### Punchlinegirl

Suppose that a=4 cm, b=21 cm, and c=28 cm. Furthermore, suppose that the electric field at a point 11 cm from the center is measured to be 2420 N/C radially inward while the electric field at a point 63 cm from the center is 130 N/C radially outward. Find the charge on the insulating sphere.
I'm sorry I can't post the picture, but it is an insulator with a radius 4 cm, that is inside of another sphere with a radius of 21 cm measured from the center, which is inside of a conductor with a radius of 28 cm measured from the center.
We did part of this problem in class and found that the $$Q_e_n_c = E_1 *4\pi* R_g^2 *E_0.$$
I'm pretty sure that E_1 = 2420 N/C and E_0= 8.85 x 10^-12, but I'm a little confused about which radius to use. I thought it would be the radius of the insulator which is .04 m, but this wasn't right. Can someone please help?

2. Feb 6, 2006

### Staff: Mentor

OK, one has 3 concentric spheres. The innermost sphere is an insulator (i.e. does not conduct, or at least not very well) which is ostensibly solid of radius 0.04 m, the outermost spherical shell of inner/outer radii of 0.21/0.28 m respectively, and a middle spherical shell of inner/outer radii of 0.04/0.21 m respectively, which is likely a dielectric material (i.e. not an insulator or conductor).

E_0= 8.85 x 10^-12 C2/N-m2 is just the electric permittivity of free space. It is important to realize the units. I prefer using $\epsilon_o$.

3. Feb 7, 2006

### Punchlinegirl

Ok I see that, but I'm a little confused about which radius to use for the question. it wants the charge on the insulating sphere. Would I just use the radius of the insulator which is .04 m?

4. Feb 7, 2006

### Staff: Mentor

You are given the field at r = 0.11 m. The equation you are using, which is from Gauss' law, will allow you to find the total charge contained within that radius. (And the only thing within that radius with any charge is the insulator, if I understand the problem correctly.)

Last edited: Feb 7, 2006
5. Feb 7, 2006

### Punchlinegirl

thank you very much.. i get it now