# Potential due to Uniform Charge Distribution (3d)

## Homework Statement ## Homework Equations

dV= integral(kdQ/dR)

## The Attempt at a Solution

So, i'm familiar with these type of problems but in 2D (like a line of uniform charge).
When the y,z component is added, i'm kinda lost.

i know dQ = p*dV= p*dx*dy*dz. (atleast i think it is).
also the dR = sqrt(x^2 + (y'-y)^2+z^2) since y is the only distance changing.
Now, i get confused as to how to put together the integral. Can someone guide me and help me out?

I am studying for an exam tomorrow :C.

#### Attachments

Homework Helper
Gold Member
## V(\vec{r})=\int \frac{k \rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\, d^3 r' ##. Which, if any, of the expressions fit this description? ## \\ ## A hint is they have reversed primes and unprimes, but that is perfectly ok to do. (Take the expression I just gave you and reverse the primes and unprimes). ## \\ ## Also, ## |\vec{r}-\vec{r}'| =\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}##. ## \\ ## Also ## \rho(\vec{r}')=\rho ##. ## \\ ## This one is pretty obvious, but I don't know that I have a simple way to teach you about the limits on triple integrals inside of 10 minutes. Which one do you think might be correct?

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## V(\vec{r})=\int \frac{k \rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\, d^3 r' ##. Which, if any, of the expressions fit this description? ## \\ ## A hint is they have reversed primes and unprimes, but that is perfectly ok to do. (Take the expression I just gave you and reverse the primes and unprimes). ## \\ ## Also, ## |\vec{r}-\vec{r}'| =\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}##. ## \\ ## Also ## \rho(\vec{r}')=\rho ##.

How do i deal with the d^3(r)? Also, is the absolute(r-r') just the sqrt(x^2+(y'-y)^2+z^2)?

Homework Helper
Gold Member
How do i deal with the d^3(r)? Also, is the absolute(r-r') just the sqrt(x^2+(y'-y)^2+z^2)?
## d^3 r=dx \, dy \, dz ##. It means the same thing. And yes, see my updated post 2. I wrote out ## |r-r' | ## for you. ## \\ ## Let ## \vec{r}=(0,y,0) ##, i.e. ## x=0 ## and ## z=0 ##, and then reverse the primes and unprimes. ## \\ ## I think you might have a pretty good idea which one is the correct answer. I can verify your answer, but as a homework helper, I am not allowed to give you the answer.

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## d^3 r=dx \, dy \, dz ##. It means the same thing. And yes, see my updated post 2. I wrote out ## |r-r' | ## for you. ## \\ ## Let ## \vec{r}=(0,y,0) ##, i.e. ## x=0 ## and ## z=0 ##, and then reverse the primes and unprimes. ## \\ ## I think you might have a pretty good idea which one is the correct answer. I can verify your answer, but as a homework helper, I am not allowed to give you the answer.

Ehh. It's not really homework. I'm doing a bunch of practice exams to prepare my exam tomorrow.

So the answer seems to be e) but is the kp/sqrt() just a constant then? and can be taken out of the integrals? The integral only depends on the limits of the box right?

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