I am trying to derive the formula for E due to an infinite sheet of charge with a charge density of $ \rho C/m^2$. I assumed the sheet is on $yz$-plane. I used Coulomb's law to get an equation and integrated the expression that over $yz$-plane. But, I have not succeeded in deriving the correct expression. The answer I am getting is $0$.

Below is the picture of my work. Kindly, have a look and let me know where did I make mistakes. In actual, E due to a charge sheet is constant and the correct expression is

E $=\rho/2\epsilon$0 aN , where aN is unit vector normal to the sheet.

enter image description here


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Method 1 (Gauss’ law):

Just simply use Gauss’ law:

$$\int_{\partial V} \vec{E} \cdot \vec{da} = \frac{Q}{\epsilon_0}.$$

A pillbox using Griffiths’ language is useful to calculate $\vec{E}$. The pillbox has some area $A$. And due to symmetry we expect the electric field to be perpendicular to the infinite sheet. Imagine putting a test charge above it, in which way does it move? Right, perpendicular to the sheet. Using $Q=\rho A$ for the charge enclosed in the pillbox we get:

$$ \rho A = \epsilon_0 \int_{\partial V} |\vec{E}| |\vec{da}| = \epsilon_0 \int_{\partial V} E da = \epsilon_0 E \int_{\partial V} da = \epsilon_0 (2AE), $$

since we expect $E$ to be constant for fixed distance for the infinite sheet. Note that the sides of the pillbox do not contribute to the integral since $\vec{E} \cdot \vec{da} = 0$ in that case.

All together we find that $E=\frac{\rho}{2 \epsilon_0}$ and the direction we thought already of is some unit vector $\hat{n}$ orthogonal to the infinite sheet:

$$ \vec{E} = \frac{\rho}{2 \epsilon_0} \hat{n} .$$

Method 2: (Coulomb/direct calculation) Another method goes as follows:

$$E=E_x= k \int \frac{x}{(r^2 + x^2)^{3/2}} r dr d\theta = 2\pi k \int \frac{xr}{(r^2 + x^2)^{3/2}} dr = 2\pi kx [ (r^2 + x^2)^{-1/2}]^0_{\infty} = 2\pi k x \frac{1}{x}= 2\pi k.$$ Let us see, I called $$k= \frac{\rho}{4 \epsilon_0 \pi}$$ we get indeed that $E=\frac{\rho}{2 \epsilon_0}$.

Errors in your calculation: - the $y$ in the nominator should be a $x$. - missing term in the denominator, namely $z^2$ because now you consider an infinite line and integrate over a surface.

  • $\begingroup$ Thanks for answering. I know that it could solved using Gauss' law. I wanted to derive this using Coulomb's law. Even for that, I have a text book at my hand in which the expression is derived using Coulomb's law. But the strategy in the book is somewhat different. I wanted to derive it with the approach I have shown above and the thing I want to know is what is wrong with my approach. $\endgroup$ – Goodfellow Aug 15 '18 at 22:45
  • $\begingroup$ Uhm okay that is a tough one... $\endgroup$ – Dani Aug 15 '18 at 22:46
  • $\begingroup$ Wouldn’t it be more easy using polar coordinates such that $x^2 + y^2 = r^2$ and $y=r sin(\theta)$ ? $\endgroup$ – Dani Aug 15 '18 at 22:49
  • $\begingroup$ I don't see why I should use polar coordinates, it is a planer sheet. Anyway, I tried that too but didn't work out. $\endgroup$ – Goodfellow Aug 15 '18 at 22:53
  • 2
    $\begingroup$ Your integral does not hold. You are missing a $z^2$ term in the square root at the beginning. I mean $x^2+y^2$ should be $x^2 + y^2 +z^2$ in $(x^2+y^2)^{-3/2}$ also.Then use $z^2 + y^2 =r^2$ to solve the integral. Then use $dA=dydz=rdrd\theta$ and integrate over these two variables. The $x$ should drop out at the end. $\endgroup$ – Dani Aug 15 '18 at 23:17

Are you looking to do the integrations by hand? Because

$r = x \hat{x} + y \hat{y} + z \hat{z}$

$r^\prime = y^\prime \hat{y} + z^\prime \hat{z}$

Should yield the correct answer, but the integrations are messy, unless you go to cylindrical coordinates

  • $\begingroup$ That is such a tedious and long method of dealing with this problem. See my added solution (method 2) how quick and easy it can be :) $\endgroup$ – Dani Aug 15 '18 at 23:46
  • $\begingroup$ You beat me to it. And not if you use mathematica :) $\endgroup$ – pmac Aug 15 '18 at 23:49
  • $\begingroup$ Actually, the integration for the y- and z-direction does take a few minutes. All that for a simple $0$. $\endgroup$ – pmac Aug 15 '18 at 23:50

Use cylindrical coordinates. The field (on axis) of a ring of charge (radius $R$, charge density $\lambda$) goes like:

$$ E(z) = \frac{1}{2\epsilon_0}\frac{ R z}{(z^2 + R^2)^{\frac 3 2}} $$

then integrate over $R$, using:

$$ \int{\frac{ R z}{(z^2 + R^2)^{\frac 3 2}dR}}=-\frac z {\sqrt{z^2+R^2}}\rightarrow 1$$

  • $\begingroup$ That is what I did in my answer... does not matter but it does not contribute something new $\endgroup$ – Dani Aug 16 '18 at 15:24
  • $\begingroup$ if that's what you did in your answer, why is your answer wrong? It's hard to read, but it looks like you're using cartesian coordinates. Cyclindrical coordinates does produce $0$. $\endgroup$ – JEB Aug 17 '18 at 2:33
  • $\begingroup$ does not produce $0$. $\endgroup$ – JEB Aug 17 '18 at 2:54

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