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The above picture requires us to find the flux through the square sheet, due to a proton. Treating the sheet as one plane of a Gaussian cube, I found the flux using, $$\text{Flux}= \frac{E*dA}{6}$$ This gives the answer $9.6 \frac{nNm^{2}}{C}$ However, the actual answer is $3.01 \frac{nNm^{2}}{C}$

Can someone please explain what I'm doing wrong? Thanks.

  • 1
    $\begingroup$ what is $dA$? Do you mean $d^2$? . $\endgroup$ – ZeroTheHero Jul 7 '17 at 11:32
  • $\begingroup$ I would guess dA refers to infinitesimal area. $\endgroup$ – Communisty Jul 7 '17 at 11:38
  • $\begingroup$ The flux should be $\int \vec{E} \cdot \mathrm{d}A$ or $E \cdot d^2 $ $\endgroup$ – Apoorv Potnis Jul 7 '17 at 12:09

If I read your question correctly then the flux through a complete box would be $\Phi_{total}=q_{encl}/\epsilon_0$ By symmetry one side would be $1/6$ of this, or $$ \Phi_{side}=q_{encl}/{6\epsilon_0}=\frac{1.6\times 10^{-19}}{6\times 8.85\times 10^{-12}}=3.01\times 10^{-9}\, . $$ The distance between the point charge and the plane or the area of your box do not enter in the calculation of $\Phi_{total}$ since the net flux is just the enclosed charged divided by $\epsilon_0$.

  • $\begingroup$ d is used in the calculation, as d^2. And (d/2)^2 is used in the calculation of E in place of r^2. $\endgroup$ – Antara Kulkarni Jul 7 '17 at 12:38
  • $\begingroup$ @AntaraKulkarni My original answer was incorrect but I've corrected it now. $\endgroup$ – ZeroTheHero Jul 7 '17 at 12:41

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