Flux through square sheet due to point charge 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.

• what is $dA$? Do you mean $d^2$? . – ZeroTheHero Jul 7 '17 at 11:32
• I would guess dA refers to infinitesimal area. – Communisty Jul 7 '17 at 11:38
• The flux should be $\int \vec{E} \cdot \mathrm{d}A$ or $E \cdot d^2$ – 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 $\epsilon_0$.