How Do You Calculate Electric Potential for a Given Charge Distribution?

[RJLiberator]

Homework Statement

The volume charge density for some charge configuration is given as $\rho (x,y,z) = \lambda \delta(x) \delta(z) [\theta(y+L)-\theta(y-L)]$ where $\theta(x)$ is the step function, defined as $\theta(x)=1$ for x>0 and 0 for x<0.

a) Calculate V(x,y,z), the potential created by this charge distribution at the position (x,y,z), using infinity as the reference point.
b) Let n denote a unit vector in the direction of the electric field created by this charge distribution at the position (x,0,z). Find n.

Homework Equations

The Attempt at a Solution

V(x,y,z) = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho}{r} d \tau

Well, from here we plug in our charge distribution. But we note that for this to make sense, x = z = 0 must be the case as the delta functions require this condition. So we observe only (0,y,0).

the 'script' r becomes only y and the derivatives just become dy.

V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0} \int \frac {[\theta(y+L)-\theta(y-L)]} {y} dy

I have no idea how to solve this integration. What have I done wrong?

 

[TSny]

You need to distinguish between the coordinates (x, y, z) of the point where you are finding V and the coordinates (x, y, z) of points within the charge distribution. Often people use primes on the coordinates of the charge distribution. So, you have $\rho(x', y', z')$ and $d\tau = dx'dy'dz'$.

 

[RJLiberator]

Ok, I did that initially. So we get a *somewhat* similar result:

<br /> V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0} \int \int \int \frac {[\theta(y+L)-\theta(y-L)]} {(x&#039;)+(y-y&#039;)+(z&#039;)} dx&#039;dy&#039;dz&#039;

Where I let x and z equal 0 due to the dirac delta functions.

I have to think that this is the correct integral, but I have no idea how to solve this or what the bounds could possibly be. How is this a linear charge density but over 3 integrations? I originally thought I had to do the x' y' and z' but figured that this must be just in y.

 

[TSny]

Ok, I did that initially. So we get a *somewhat* similar result:

<br /> V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0} \int \int \int \frac {[\theta(y+L)-\theta(y-L)]} {(x&#039;)+(y-y&#039;)+(z&#039;)} dx&#039;dy&#039;dz&#039;

Where I let x and z equal 0 due to the dirac delta functions.

Note $\rho (x',y',z') = \lambda \delta(x') \delta(z') [\theta(y'+L)-\theta(y'-L)]$. Note the primes on the coordinates on the right hand side.

 

[RJLiberator]

Ah, that is an error on my part that could lead me to the correct answer.

<br /> V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0} \int \int \int \frac {[\theta(y&#039;+L)-\theta(y&#039;-L)]} {(x-x&#039;)+(y-y&#039;)+(z-z&#039;)} dx&#039;dy&#039;dz&#039;

After letting x' = 0 and z' = 0 due to the dirac delta functions we get:

<br /> V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0} \int \int \int \frac {[\theta(y&#039;+L)-\theta(y&#039;-L)]} {(x)+(y-y&#039;)+(z)} dx&#039;dy&#039;dz&#039;

Which is nicer, but still does not solve my problem with knowing how to integrate over the step function. The dx' and the dz' become easier to solve, but the dy' is a nuisance.

 

[TSny]

You are not expressing the distance $r$ correctly.

The delta functions $\delta(x')$ and $\delta(z')$ are used to simplify the integrations over $x'$ and $z'$. After using the delta functions there will be no integration left over $x'$ or $z'$.

 

[RJLiberator]

Ok, but even so -- when I look at this problem, I don't know how to evaluate the step function in the integral. I see how the dirac delta functions make the integration over x' and z' trivial, but dy' over the step function is a real nuisance. No?

 

[TSny]

To handle the step functions, consider the value of $\theta (y' + L) - \theta (y' - L)$ for the three regions:

(i) $\,\,\,y' < -L$
(ii) $\,\,\,-L < y' < L$
(iii) $\,\,\,y' > L$

 

[RJLiberator]

Okay, that helps out a lot. I've now split the integration up into three regions as stated.

If we observe region I, then we know that the step function turns out to be 0.
In region III, we know the step function is (1-1) = 0 so the function turns out to be 0.
In region II, we are interested in the results.

The problem is that the function results in a value of 0 for both parts. If the teacher did not specify what to do at the part theta(0), can I continue with it?

 

[TSny]

Yes, $\theta (y' + L) - \theta (y' - L) = 0$ in the regions $y' < -L$ and $y' > L$. So, the integrand is zero in these two regions.

What is the value of $\theta (y' + L) - \theta (y' - L) = 0$ in the region $-L < y' < L$?

I'm not understanding this:

The problem is that the function results in a value of 0 for both parts. If the teacher did not specify what to do at the part theta(0), can I continue with it?

 

[RJLiberator]

My question is, evaluation $\theta(0)$ seems to be undefined by the given question. How can I proceed? Since hte integration makes L-L in the step function.

 

[TSny]

You don't need to worry about the the value of $\theta(x)$ at $x = 0$. When you integrate a piecewise continuous function, such as $\theta(x)$, the value of the integral does not depend on the value of the function at the point(s) of discontinuity. Whether you define $\theta(0) = 0$, or $\theta(0) = 1$, or let $\theta(0)$ be any other finite value, the integral will have the same value. Changing the integrand at only one point of the domain of integration does not affect the integral. ("No area under a point.")

 

[RJLiberator]

Hm. Interesting.

So my integration is set up as:

V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0}\int_{-L}^{L} \frac {[\theta(y&#039;+L)-\theta(y&#039;-L)]} {(x&#039;)+(y-y&#039;)+(z&#039;)}dy&#039;

But I'm still stuck solving an integration over this step function. Let's say I take your knowledge from the last post, let's see what I can do with that.

If I say the step function will be "1" somewhere under that curve, then i can say

V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0}\int_{-L}^{L} \frac {1} {(x&#039;)+(y-y&#039;)+(z&#039;)}dy&#039;

Correct? And this is easy to evaluate.

 

[TSny]

Right. You just need to integrate between $-L$ and $L$ and in this region you can replace $\theta(y'+L) - \theta (y'-L)$ by $1$. You don't need to worry about the value of the theta functions at $y' = L$ or $y' = -L$.

But you still don't have the correct expression for the denominator.

 

[RJLiberator]

The x and z must leave the denominator leaving just $\frac{1}{(y-y')}$, correct? But why would this be so?

I understand the step-function idea now over integration, thanks to your help. That was a big issue I had on this problem.

 

[TSny]

The x' and z' must leave the denominator leaving just $\frac{1}{(y-y')}$, correct?

No. $x'$ and $z'$ will be gone, but not $x$ and $z$.

Consider two points in the x-y plane: $(x_1, y_1)$ and $(x_2, y_2)$. How do you find the distance $r$ between the two points?

 

[RJLiberator]

Ah...
I see now.

It needs to be
\sqrt{x^2+(y-y&#039;)^2+z^2}

 

[TSny]

Yes.

 

[RJLiberator]

Thank you for the help with this problem.

 

[RJLiberator]

However, the integral seems to be extremely difficult. How do I work with this thing??

V(x,y,z) = \frac{\lambda}{4 \pi \epsilon_0}\int_{-L}^{L} \frac {1} {\sqrt{x^2+(y-y&#039;)^2+z^2}}dy&#039;

 

[TSny]

Can be done with a trig substitution. Or try http://integral-table.com/