Question

A charge $Q$ is present inside a cube, then find out flux due to cube and flux due to each face of a cube?

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Solution

Step 1: Given dataCharge inside the cube = $Q$The dielectric constant of the vacuum = ${\epsilon }_{0}$Faces of a cube = $6$Step 2: Concept appliedGauss's lawGauss's law for electricity states that the electric flux $\varphi$ across any closed surface is proportional to the net electric charge $q$enclosed by the surface.i.e., $\varphi =\frac{q}{{\epsilon }_{0}}$Flux ($\varphi$) is defined as the electric field flowing out through a specific area.Consider a surface element $d\stackrel{\to }{S}=\stackrel{^}{n}dS$ in an electric field $\stackrel{\to }{E}$ , where $\stackrel{^}{n}$ is the outward unit vector normal to the surface element. The quantity $d\varphi =\stackrel{\to }{E}.d\stackrel{\to }{S}=E\mathrm{cos}\theta dS$ is called the flux of $\stackrel{\to }{E}$ through $d\stackrel{\to }{S}$. If we consider that $\frac{q}{{\epsilon }_{0}}$ number of lines of force emanate isotropically from a point charge $q$ then the idea of electric flux becomes meaningful. If we use this picture of an electric field then $d\varphi =\stackrel{\to }{E}.d\stackrel{\to }{S}$ becomes equal to the number of lines of force passing through the area $d\stackrel{\to }{S}$.The flux of $\stackrel{\to }{E}$ over any arbitrary surface $S$ is given by the integral $\Phi =\int d\varphi ={\int }_{S}\stackrel{\to }{E}.d\stackrel{\to }{S}$.Step 3: Calculation and conclusionSince Charge $Q$ is present inside the cube, the flux through the cube will be$\varphi =\frac{q}{{\epsilon }_{0}}$The flux will be one-sixth of the total flux=$1}{6}$(since a cube has 6 faces) Flux through each face will be = $\frac{q}{6{\epsilon }_{0}}$Hence, flux due to cube is $\varphi =\frac{q}{{\epsilon }_{0}}$ and flux due to each face of a cube is $\frac{q}{6{\epsilon }_{0}}$

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