A point charge of 2.0 ; mu C is at the centre | vedclass

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(D) According to Gauss's Law,the net electric flux $(\phi_{	ext{Net}})$ through any closed surface is equal to the total charge enclosed $(q)$ divide

Vedclass · Accepted Answer

$2.26 	imes 10^{5} \; N \; m^{2} C^{-1}$

Vedclass · Answer

(D) According to Gauss's Law,the net electric flux $(\phi_{	ext{Net}})$ through any closed surface is equal to the total charge enclosed $(q)$ divided by the permittivity of free space $(\varepsilon_{0})$.$\phi_{	ext{Net}} = \frac{q}{\varepsilon_{0}}$Given:Charge $q = 2.0 \; \mu C = 2.0 	imes 10^{-6} \; C$$\varepsilon_{0} = 8.854 	imes 10^{-12} \; N^{-1} m^{-2} C^{2}$Substituting the values:$\phi_{	ext{Net}} = \frac{2.0 	imes 10^{-6}}{8.854 	imes 10^{-12}}$$\phi_{	ext{Net}} \approx 0.22588 	imes 10^{6} \; N \; m^{2} C^{-1}$$\phi_{	ext{Net}} \approx 2.26 	imes 10^{5} \; N \; m^{2} C^{-1}$Thus,the net electric flux through the surface is $2.26 	imes 10^{5} \; N \; m^{2} C^{-1}$.

$A$ point charge of $2.0 \; \mu C$ is at the centre of a cubic Gaussian surface $9.0 \; cm$ on edge. What is the net electric flux through the surface?

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