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

by KaiserBrandon
Tags: electromagnetism, gauss law, sphere, uniform charge
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KaiserBrandon
#1
Oct6-10, 07:30 PM
P: 54
1. The problem statement, all variables and given/known data
Two spheres, each of radius R and carrying uniform charge densities +[tex]\rho[/tex]
and [tex]-\rho[/tex], respectively, are placed so that they partially overlap.
Call the vector from the positive centre to the negative centre [tex]\vec{d}[/tex]. Show
that the field in the region of overlap is constant and find its value. Use
Gaussís 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
[tex](r\rho)/(3\epsilon)[/tex]
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 [tex]\vec{r}[/tex]. And from P to the centre of the negative sphere, I denoted [tex]\vec{r'}[/tex]. so [tex]\vec{r'}=\vec{d}-\vec{r}[/tex]. So in order for P to be inside the spheres, [tex]|\vec{r}|<R[/tex] and [tex]|\vec{d}-\vec{r}|<R[/tex]. So using the law of superposition, inside the overlap, the electric is
[tex]E = (|\vec{r}|-|\vec{d}-\vec{r}|)\rho/3\epsilon[/tex]
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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granpa
#2
Oct6-10, 09:46 PM
P: 2,258
the electric field is a vector so why on earth are you reducing r and d-r to scalars?
KaiserBrandon
#3
Oct7-10, 12:23 PM
P: 54
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.

KaiserBrandon
#4
Oct7-10, 01:45 PM
P: 54
Electric field in the overlap of two solid, uniformly charged spheres

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

[tex]\vec{E}=\frac{\rho*d}{3*\epsilon}*\hat{i}[/tex]

where d is the magnitude of [tex]\vec{d}[/tex]

And this is under the condition that [tex]\vec{d}[/tex] runs along the x axis.
granpa
#5
Oct7-10, 06:57 PM
P: 2,258
sometimes you just need to sleep on it and get a fresh perpective on it in the morning


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