Five point charges frac{1}{pi}, frac{2}{pi}, | vedclass

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(A) According to Gauss's law,the total electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.Here,th

Vedclass · Accepted Answer

$180 \ Nm^2 C^{-1}$

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(A) According to Gauss's law,the total electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.Here,the total enclosed charge $q_{enclosed} = (\frac{1}{\pi} + \frac{2}{\pi} + \frac{3}{\pi} + \frac{4}{\pi} - \frac{5}{\pi}) 	imes 10^{-9} \ C = \frac{5}{\pi} 	imes 10^{-9} \ C$.Substituting this into Gauss's law:$\phi = \frac{5 	imes 10^{-9}}{\pi \varepsilon_0}$.We know that $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 \ Nm^2 C^{-2}$,so $\frac{1}{\pi \varepsilon_0} = 4 	imes 9 	imes 10^9 = 36 	imes 10^9$.Therefore,$\phi = 5 	imes 10^{-9} 	imes 36 	imes 10^9 = 180 \ Nm^2 C^{-1}$.

Five point charges $\frac{1}{\pi}, \frac{2}{\pi}, \frac{3}{\pi}, \frac{4}{\pi}$ and $\frac{-5}{\pi} \ nC$ are located inside a pyramid. The total electric flux through the surface of the pyramid is

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