Arrangements of charges are shown in the figu | vedclass

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(C) According to Gauss's Law,the electric flux $\phi$ linked with a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$,where $q_{e

Vedclass · Accepted Answer

$\frac{-q}{\varepsilon_0}, \frac{q}{\varepsilon_0}$

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(C) According to Gauss's Law,the electric flux $\phi$ linked with a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$,where $q_{enclosed}$ is the net charge enclosed by the surface.For surface $P$:The charges enclosed are $+q, -q, -q$. Net charge $q_{P} = (+q) + (-q) + (-q) = -q$.Therefore,flux $\phi_{P} = \frac{-q}{\varepsilon_0}$.For surface $Q$:The charges enclosed are $+q, -q, +q$. Net charge $q_{Q} = (+q) + (-q) + (+q) = +q$.Therefore,flux $\phi_{Q} = \frac{q}{\varepsilon_0}$.Thus,the flux linked with $P$ and $Q$ are $\frac{-q}{\varepsilon_0}$ and $\frac{q}{\varepsilon_0}$ respectively.

Arrangements of charges are shown in the figure. Flux linked with the closed surfaces $P$ and $Q$ respectively are . . . . . . and . . . . . . .

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