So if we have a large sheet that is not uniformly charged and is NOT a conductor, how can I find an expression for the electric field everywhere?

Things we know about the sheet:

  • the width is 2b

  • it is lying on the yz plane (centered at the origin)

  • the charge density varies as ρ = ρ(0)(x/b)^2

  • ρ(0) is constant

I'm not sure if I can actually use Gauss' law because there's not the same symmetry as if it was uniformly charged. Is there some other law of electricity or formula that exists? How do I go about tackling this problem?


closed as off-topic by Kyle Kanos, ACuriousMind, HDE 226868, Neuneck, Martin Aug 3 '15 at 12:16

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    $\begingroup$ Welcome to Physics! Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. $\endgroup$ – Kyle Kanos Aug 1 '15 at 20:44
  • $\begingroup$ By 'width' I assume you mean the thickness of the sheet? In that case the situation is still very symmetric and uniform. For points outside the sheet you can just treat the sheet as infinitely thin and use Gauss's law. And for points that are inside you treat the sheet as two separate infinitely thin sheets placed on both sides of that point. $\endgroup$ – SpiderPig Aug 2 '15 at 6:40

If you have some (static) charge distribution $\rho(\mathbf{x})$ the the electric field is given by:

$$ \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} $$

or it might be easier to calculate the potential using Poisson's equation:

$$ \nabla^2 V = -\frac{\rho}{\varepsilon_0} $$

and then calculate the field using:

$$ \mathbf{E} = -\nabla V $$

In your case you are given $\rho$ so the equations are easy to set up. Solving them might be harder ...

  • $\begingroup$ I'm a little surprised to be downvoted. This seemed to me a reasonable response to a question that borders on homework. Have I made a factual error, or is the downvote a philosophical objection? $\endgroup$ – John Rennie Aug 2 '15 at 5:54

Here's the easy way to do it.

Do Gauss's law for a single uniformly charged $yz$-plane with surface charge $\sigma$. You'll find that the electric field is distance-independent except for its sign; $$\vec E = \frac{\sigma}{2\epsilon_0} ~ \operatorname{sgn}(x) ~ \hat x.$$Here $\hat x$ is the unit vector in the $x$-direction and $\operatorname{sgn}(x) = x / |x|$ is the "sign function" which is $+1$ for positive $x$, $-1$ for negative $x$, and $0$ when $x = 0$.

Now use the principle of superposition: replace $\sigma$ with an infinitesimal charge $\rho(x')~dx'$, and shift the center of the $\operatorname{sgn}(x)$ term to $x - x'$ to reflect its new center. You get from their superposition:$$\vec E(x) = \frac{\rho_0}{2\epsilon} ~ \hat x ~ \int_{-b}^{b}dx'~\left(\frac{x'}{b}\right)^2~\operatorname{sgn}(x - x')$$ You should be able to handle this integral the same way you handle any integral with an absolute value: break it up into two parts, $x' > x$ and $x' < x$, and "fill in" the appropriate $\text{sgn}$ for each. You don't have to do this if $x > b$ or $x < -b$ of course, so you should get one answer for $|x| > b$ corresponding to all of the charge being "shrunk down" into the equivalent surface charge $\sigma$, and another answer where you're inside the material and the charges on either side of you are opposing each other.


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