q_1, q_2, q_3 and q_4 are point charges locat | vedclass

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(B) According to Gauss's law,the total electric flux $\Phi_E$ through a closed surface is equal to the total enclosed charge divided by the permittivi

Vedclass · Accepted Answer

$\oint_S (\vec{E}_1 + \vec{E}_2 + \vec{E}_3) \cdot d\vec{A} = \frac{q_1 + q_2 + q_3}{\varepsilon_0}$

Vedclass · Answer

(B) According to Gauss's law,the total electric flux $\Phi_E$ through a closed surface is equal to the total enclosed charge divided by the permittivity of free space $\varepsilon_0$.$\Phi_E = \oint_S \vec{E}_{net} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}$In the given figure,the charges enclosed by the Gaussian surface $S$ are $q_1, q_2$ and $q_3$. The charge $q_4$ is outside the surface.Therefore,the net electric field $\vec{E}_{net}$ at any point on the surface is the vector sum of the electric fields produced by all charges (both inside and outside the surface),i.e.,$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \vec{E}_4$.However,the flux through the surface depends only on the enclosed charge $Q_{enc} = q_1 + q_2 + q_3$.Thus,$\oint_S (\vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \vec{E}_4) \cdot d\vec{A} = \frac{q_1 + q_2 + q_3}{\varepsilon_0}$.Since the expression in option $(b)$ represents the flux due to the enclosed charges,and the integral of the net field over the surface is equal to the flux,option $(b)$ is the correct representation of the law applied to this system.

$q_1, q_2, q_3$ and $q_4$ are point charges located at points as shown in the figure and $S$ is a spherical Gaussian surface of radius $R$. Which of the following is true according to Gauss's law?

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