# Homework Help: Magnitude of an Electric field on a point by 2 charged spheres

1. Jan 24, 2010

### KayleighK

1. The problem statement, all variables and given/known data

The left-hand sphere has a positive charge Q and the right-hand sphere has a negative charge -Q . Charge is distibuted uniformly over each of two spherical volumes with radius R. One sphere of charge is centered at the origin and the other at x=2R .
Find the magnitude of the net electric field at the point R/2 on the x-axis

2. Relevant equations

E=$$\frac{1}{4\pi\epsilon_{0}}$$ ($$\frac{Q}{R^{2}}$$

3. The attempt at a solution

Since the point is located within the first sphere, I thought the electric field would be zero.
Then I typed in:

E=$$\frac{1}{4\pi\epsilon_{0}}$$ ($$\frac{Q}{\frac{3}{2}R^{2}}$$

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 24, 2010

### rl.bhat

Since the point is located within the first sphere, I thought the electric field would be zero.

But the sphere is not a conductor. So the field inside a uniformly charge sphere the is not equal to zero.

3. Jan 24, 2010

### KayleighK

Oh ok, so then the E field for the first sphere would be:

$$\frac{1}{4_{0}\epsilon\pi}$$$$\frac{Q}{\frac{1}{2}R^{2}}$$

And then since both fields of the spheres are pointing the same direction I will just add the E fields of the first sphere with the second sphere?

4. Jan 24, 2010

### ideasrule

Why? The shell theorem states that the part of the sphere from r=1/2R to R creates no net electric field. The sphere "under" that shell has radius 1/2R, so what charge must it have?

5. Jan 24, 2010

### KayleighK

Would it have half the charge it originally had? Q/2?

6. Jan 24, 2010

### ideasrule

Would it? How much of the original volume does the smaller sphere have?

7. Jan 24, 2010

### KayleighK

It would have 1/8 less volume...so then Q/8

8. Jan 24, 2010

### KayleighK

Ok, I finally understand the problem now. Thank you so much for your help!!!! I appreciate it =)