# Electric field in the overlap of two solid, uniformly charged spheres

1. Oct 6, 2010

### KaiserBrandon

1. The problem statement, all variables and given/known data
Two spheres, each of radius R and carrying uniform charge densities +$$\rho$$
and $$-\rho$$, respectively, are placed so that they partially overlap.
Call the vector from the positive centre to the negative centre $$\vec{d}$$. Show
that the field in the region of overlap is constant and find its value. Use
Gauss’s law to find the electric field inside a uniformly charged sphere
first.

2. Relevant equations
law of superposition
Gauss Law

3. The attempt at a solution
I found the field inside one sphere to be
$$(r\rho)/(3\epsilon)$$
in the radial direction. Now for the overlapping spheres, I said that the vector from the centre of the positive sphere to some point P in the interlapping area is $$\vec{r}$$. And from P to the centre of the negative sphere, I denoted $$\vec{r'}$$. so $$\vec{r'}=\vec{d}-\vec{r}$$. So in order for P to be inside the spheres, $$|\vec{r}|<R$$ and $$|\vec{d}-\vec{r}|<R$$. So using the law of superposition, inside the overlap, the electric is
$$E = (|\vec{r}|-|\vec{d}-\vec{r}|)\rho/3\epsilon$$
in the radial direction, with the boundaries in effect. Now I am stumped here, as I'm unsure how to reduce this to a constant. Any suggestions?

2. Oct 6, 2010

### granpa

the electric field is a vector so why on earth are you reducing r and d-r to scalars?

3. Oct 7, 2010

### KaiserBrandon

yep, realized my mistake while sitting in my thermodynamics class this morning. It's funny how I usually figure stuff out while I'm not actually trying to do the question.

4. Oct 7, 2010

### KaiserBrandon

k, so I changed the E function to Cartesian coordinates. So in the overlap I got:

$$\vec{E}=\frac{\rho*d}{3*\epsilon}*\hat{i}$$

where d is the magnitude of $$\vec{d}$$

And this is under the condition that $$\vec{d}$$ runs along the x axis.

5. Oct 7, 2010

### granpa

sometimes you just need to sleep on it and get a fresh perpective on it in the morning