electric field in the overlap of two solid, uniformly charged spheresby KaiserBrandon Tags: electromagnetism, gauss law, sphere, uniform charge 

#1
Oct610, 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 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 [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? 



#2
Oct610, 09:46 PM

P: 2,258

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




#3
Oct710, 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.




#4
Oct710, 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. 



#5
Oct710, 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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