If electric flux through a cubic Gaussian sur | vedclass

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Question

(A) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q}{\varepsilon_0}$,where $q$ is the enclosed

Vedclass · Accepted Answer

$2 \mu 	ext{C}$

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(A) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q}{\varepsilon_0}$,where $q$ is the enclosed charge and $\varepsilon_0$ is the permittivity of free space $(8.854 	imes 10^{-12} 	ext{ C}^2 	ext{N}^{-1} 	ext{m}^{-2})$.Given $\phi = 1.9 	imes 10^5 	ext{ Nm}^2 	ext{C}^{-1}$.Rearranging the formula to find the charge: $q = \phi \varepsilon_0$.Substituting the values: $q = (1.9 	imes 10^5) 	imes (8.854 	imes 10^{-12})$.$q = 16.8226 	imes 10^{-7} 	ext{ C}$.$q \approx 1.68 	imes 10^{-6} 	ext{ C} = 1.68 \mu 	ext{C}$.Rounding to the nearest provided option,we get $q \approx 2.0 \mu 	ext{C}$.

If electric flux through a cubic Gaussian surface is $1.9 \times 10^5 \text{ Nm}^2 \text{C}^{-1}$,then the electric charge at its center is . . . . . . . (Edge length of cube = $9.0 \text{ cm}$).

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