Semi-infinite slab of charges -- find E and charge density

[allison west]

Homework Statement

A semi-infinite (infinite in y and z, bounded in x) slab of charges carries a charge per unit volume $\rho$. Electric potential due to this slab is a function of horizontal distance, x from the center of the slab. It is linear for $x \lt -1m$ & $x \gt 1m$, and is given by $V(x) = \frac 15 2 x^2 - \frac 25 2$ for $-1m \leq x \leq 1m$. Find the x-component of the electric field. Use Gauss' law to find the charge density $\rho$ of the slab.

Vector Potential Plot:

Homework Equations

Gauss' Law:

$\frac {q_{encl}} {\epsilon _0} = \oint \vec E \cdot d \vec A$

$\rho = \frac Q V$

$\vec E = -\nabla V$

The Attempt at a Solution

Using the equation for the electric potential from the plot I have $V(x) = 15x + V_0$ for $x$ being outside of the bounds. ($x \lt -1m$ and $x \gt 1m$)
I get the electric field to be:

$\vec E = -\nabla V$

$\vec E = -15 \hat x$

Then for the electric potential outside of the bounds, $V(x) = \frac 15 2 x^2 - \frac 25 2$
Here I get the electric field to be:

$\vec E = -15x \hat x$

But I'm not sure that it makes sense to have two different values for the electric field? Shouldn't I just get one function?

Then when I solve for charge density I get:

$\rho = \frac {q_{encl}} V$

$Volume = A x$

$\rho = \frac {[\oint \vec E \cdot d \vec A] \epsilon _0 } {A x}$

$\rho = \frac {\epsilon _0} x E_x$

Which doesn't really make sense because I get the charge density falling off at $\frac 1 x$ outside of the bounds, and I get it to be constant within the slab. Which should be the other way around (it should be constant outside of the bounds of $x$). I'm not sure if I attempted this problem wrong, any help would be appreciated, thanks!​

 

[TSny]

Hello

$V(x) = \frac 15 2 x^2 - \frac 25 2$ for $-1m \leq x \leq 1m$.

Those numerical coefficients look a little odd to me. Is there a typographical error here? Are they supposed to be $\frac {15}{2}$ and $\frac {25}{2}$?

Vector Potential Plot:

Did you mean "Scalar potential plot"?

3. The Attempt at a Solution

Using the equation for the electric potential from the plot I have $V(x) = 15x + V_0$ for $x$ being outside of the bounds. ($x \lt -1m$ and $x \gt 1m$)

Can you explain how you got this result for $V(x)$?​

Then for the electric potential outside of the bounds, $V(x) = \frac 15 2 x^2 - \frac 25 2$

Did you mean inside the bounds?

But I'm not sure that it makes sense to have two different values for the electric field? Shouldn't I just get one function?

The expression for the field outside the bounds can be different than the expression for the field inside the bounds.

$\rho = \frac {[\oint \vec E \cdot d \vec A] \epsilon _0 } {A x}$

$\rho = \frac {\epsilon _0} x E_x$

Did you evaluate the surface integral correctly if E is proportional to x?
​

 

[allison west]

Hello
Those numerical coefficients look a little odd to me. Is there a typographical error here? Are they supposed to be $\frac {15}{2}$ and $\frac {25}{2}$?

Did you mean "Scalar potential plot"?

Can you explain how you got this result for $V(x)$?​

Did you mean inside the bounds?

The expression for the field outside the bounds can be different than the expression for the field inside the bounds.

Did you evaluate the surface integral correctly if E is proportional to x?
​

Yes sorry, $V(x) = \frac {15}{2} x - \frac {25}{2}$

Yes, I think it would be scalar potential.

I got $V(x)$ from the plot which I guessed given the coordinates of a couple points and the fact that I knew it was linear. It looks like one point on the graph is (10, 2) and that the y-intercept is around -20 when the linear graph is extended downwards. I'm not sure how else I would find the electric potential outside of the bounds as all I know is that it is linear.

Yes, the second V(x) is for inside the bounds, sorry.

I evaluated the surface integral by using a Gaussian pillbox and taking E and A out of the integral with the direction being in the $\hat x$ direction because this is where the flux would emerge out of the slab I believe. I'm not sure what you mean by evaluating it proportional to x?

Thanks for the reply!

 

[TSny]

To find the expression for $V(x)$ for $x > 1$ m, use the fact that the potential is given to be a linear function of $x$ in this region and $V(x)$ and its first derivative must be continuous at $x = 1$ m.

 

[TSny]

I evaluated the surface integral by using a Gaussian pillbox and taking E and A out of the integral with the direction being in the $\hat x$ direction because this is where the flux would emerge out of the slab I believe. I'm not sure what you mean by evaluating it proportional to x?

You found $E_x = -15x$ for inside the bounds. I believe this is correct. Did you overlook the $x$ in $-15x$ when using $E$ in Gauss' law?

 

[allison west]

To find the expression for $V(x)$ for $x > 1$ m, use the fact that the potential is given to be a linear function of $x$ in this region and $V(x)$ and its first derivative must be continuous at $x = 1$ m.

So would it be correct to say that $V(x) = -15 x$ outside of the bounds?

When I input the values of $E_x$ I would get:

outside of the bounds:
$\rho = \frac {-15 \epsilon _0} {x}$

inside of the bounds:
$\rho = -15 {\epsilon _0}$

Or do I have to differentiate $E_x$ in Gauss' law?

 

[TSny]

So would it be correct to say that $V(x) = -15 x$ outside of the bounds?

This is not quite correct. Your original answer of $V(x) = 15 x + V_0$ is correct for $x > 1$. But I could not see how you got this result. You can determine the value of $V_0$ by choosing $V_0$ such that $V(x)$ is continuous at $x = 1$.

The problem only asks you to find $\vec E$, so you don't need to actually determine the value of $V_0$. In your first post I believe you stated that $\vec E = -15 \hat i$ for the outer region. This is correct for the part of the outer region where $x > 1$, but not for the part of the outer region where $x < -1$.

For $x < -1$, the potential is not given by $V(x) = 15 x + V_0$. A small modification is needed in the expression.

When I input the values of $E_x$ I would get:

outside of the bounds:
$\rho = \frac {-15 \epsilon _0} {x}$

This is not correct for $\rho$ in the outer region. Since there is no material in the outer region, what would you expect for $\rho$ in the outer region?

inside of the bounds:
$\rho = -15 {\epsilon _0}$

This is correct, but I don't see how you got this answer.

 

[allison west]

This is not quite correct. Your original answer of $V(x) = 15 x + V_0$ is correct for $x > 1$. But I could not see how you got this result. You can determine the value of $V_0$ by choosing $V_0$ such that $V(x)$ is continuous at $x = 1$.

The problem only asks you to find $\vec E$, so you don't need to actually determine the value of $V_0$. In your first post I believe you stated that $\vec E = -15 \hat i$ for the outer region. This is correct for the part of the outer region where $x > 1$, but not for the part of the outer region where $x < -1$.

For $x < -1$, the potential is not given by $V(x) = 15 x + V_0$. A small modification is needed in the expression.

This is not correct for $\rho$ in the outer region. Since there is no material in the outer region, what would you expect for $\rho$ in the outer region?

This is correct, but I don't see how you got this answer.

Okay, so $V(x) = -15x + V_0$ when $x \lt -1$ making $\vec E_x = 15 \hat x$.

$\rho$ in the outer region would be zero because it's outside of the slab.

I got $\rho$ inside the region by using Gauss' Law:

$\rho = \frac Q V$

$\frac {q_{encl}} {\epsilon _0} = \oint \vec E \cdot d \vec A$

$q_{encl} = E A \epsilon _0$

$V = A x$

$\rho = \frac {\epsilon _0} {x} E$

Where $E = E_x$

$\rho = -15 \epsilon _0$

Thanks for the help!

 

[TSny]

OK. But note that in your application of Gauss' law, you assumed that the charge density inside the slab is uniform. Can you justify that?

Have you studied the differential form of Gauss' law in terms of the divergence of $\vec E$?
$\vec \nabla \cdot \vec E = \rho / \epsilon_0$