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

AI Thread Summary
The discussion centers on calculating the electric field in the overlapping region of two uniformly charged spheres, one positively charged and the other negatively charged. Using Gauss's law, the electric field inside a uniformly charged sphere is established as (rρ)/(3ε). The participant initially struggled with the vector representation of the electric field but later corrected their approach by converting to Cartesian coordinates. Ultimately, they derived the electric field in the overlap as E = (ρd/3ε) in the direction of the x-axis, confirming that the field is constant in this region. The resolution emphasizes the importance of taking a break to gain clarity on complex problems.
KaiserBrandon
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Homework Statement


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.


Homework Equations


law of superposition
Gauss Law

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?
 
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the electric field is a vector so why on Earth are you reducing r and d-r to scalars?
 
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.
 
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.
 
sometimes you just need to sleep on it and get a fresh perpective on it in the morning
 
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