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What would be the total electric flux $\Phi_E$ through an infinite plane due to a point charge $q$ at a distance $d$ from the plane?

I think it should be ${q/2\epsilon_0}$ but I cannot justify that. My teacher says the flux is infinite due to the infinite area of the sheet which I cannot believe.

Qmechanic
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Sillysack Buttowski
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  • Then why do you think that it should be $q/2\epsilon_{0}$ if you cannot justify it? – Nemo Nov 06 '17 at 14:30
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    Well, this may not be proper but I know that from Gauss's law the maximum flux through any closed surface due to a charge $q$ can be $q/\epsilon_0$. Since this is an open infinite surface, I assume that half of all the field lines pass through the plane while the other half don't. Hence my conclusion of $q/2\epsilon_0$. – Sillysack Buttowski Nov 06 '17 at 14:37
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    Good. You're right. – Nemo Nov 06 '17 at 14:41
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    Your intuition is partly correct. The actual reason will be clear if you place another infinite plate at a distance $d$ from the plate and $\frac{d}{2}$ from the test charge. Now this can be thought as a closed volume and by symmetry, the flux must be distributed $q/2\epsilon _{0} $ on each plate. – physicscircus Nov 06 '17 at 14:42
  • Could you draw this? I don't really understand what you mean. – Sillysack Buttowski Nov 06 '17 at 14:57

3 Answers3

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In general the flux through an oriented open or closed surface $\:\mathrm{S}\:$ due to a point charge $\:Q\:$ is
\begin{equation} \Phi_{\mathrm{S}}=\dfrac{\Theta}{4\pi}\dfrac{Q}{\epsilon_{0}} \tag{01} \end{equation} where $\:\Theta\:$ the solid angle by which the charge $\:Q\:$ sees the surface. In our case this solid angle is half the complete $\:4\pi\:$ solid angle, that is $\:2\pi\:$, so
\begin{equation} \Phi_{\mathrm{S}}=\dfrac{2\pi}{4\pi}\dfrac{Q}{\epsilon_{0}}=\frac12\dfrac{Q}{\epsilon_{0}} \tag{02} \end{equation}

You can find special cases for the solid angles by which a point sees rectangular parallelograms in my answer therein :What is the electric field flux through the base of a cube from a point charge infinitesimally close to a vertex?.

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Frobenius
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The infinite area is a red herring. The electric flux from a point charge does not measure area, because of the inverse-square dependence of the electric field itself; instead, it measures solid angle (a well-known standard fact of electromagnetism), and this is bounded above by $4\pi$, so no regular surface can accumulate infinite flux from a point charge.

If you want, you can show this explicitly through direct integration: putting the charge at $(0,0,d)$ and the plane in the $xy$ plane integrated through polar coordinates, the flux is given by \begin{align} \Phi & = \iint \mathbf E(\mathbf r)\cdot\hat{\mathbf z}\:\mathrm dS \\ & = \int_0^\infty\int_0^{2\pi} \frac{q}{4\pi\epsilon_0}\frac{r\hat{\mathbf r}-d\hat{\mathbf z}}{(r^2+d^2)^{3/2}}\cdot\hat{\mathbf z}\:r\:\mathrm d\theta \:\mathrm dr \\ & = -\frac{qd}{4\pi\epsilon_0}\int_0^\infty\frac{r}{(r^2+d^2)^{3/2}}\mathrm dr , \end{align} which is easily seen to converge (and which can, moreover, be integrated explicitly with the substitution $r=d\tan(\phi)$ to give the primitive $d/(r^2+d^2)^{1/2}$).

Emilio Pisanty
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I think its answer is $q/\epsilon_0$ where $\epsilon_0$ is permittivity of free space. Because when flux through gaussian surface enclosing charge $q$ is $q/\epsilon_0$ and flux through any body near to this charge like plane in this case will be of course $q/\epsilon_0$.

Thomas Fritsch
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Sumaira ashraf
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  • Your answer is incorrect because the plane is not a gaussian surface enclosing the charge $q$. You will understand this looking in the Figure titled "Solid angles" in my answer. From the $\Theta=2\pi$ example the point $Q$ "sees" the "south" hemisphere and the plane underneath it by the same solid angle. And this solid angle is $\Theta=2\pi$ – Frobenius Aug 08 '19 at 23:13