A cubical Gaussian surfaces encloses 30 C per unit permittivity of charge- The electric flux through each face of the cube is10 C5 C30 C15 C

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Given cube encloses 30C-x3F5-owe know gauss law -x222E-E-x22C5-da-qo-x3F5-o-x2234- Total flux through all 6 faces -30C -xA0-per unit permittivity-x21D2- Flux through each face -306-x21D2-5C-x3F5-o

A cubical Gaussian surfaces encloses $30 C$ per unit permittivity of charge. The electric flux through each face of the cube is

$10 C$
$5 C$
$30 C$
$15 C$

Given cube encloses $\frac{30 C}{ϵ_{o}}$
we know gauss law $\oint E \cdot d a = \frac{q_{o}}{ϵ_{o}}$
$∴$ Total flux through all 6 faces $= 30 C$ per unit permittivity
$\Rightarrow$ Flux through each face $= \frac{30}{6}$

$\Rightarrow \frac{5 C}{ϵ_{o}}$

A cubical Gaussian surfaces encloses 30 C per unit permittivity of charge. The electric flux through each face of the cube is10 C5 C30 C15 C

Given cube encloses 30Cϵowe know gauss law ∮E⋅da=qoϵo∴ Total flux through all 6 faces =30C per unit permittivity⇒ Flux through each face =306⇒5Cϵo

A cubical Gaussian surfaces encloses $30 C$ per unit permittivity of charge. The electric flux through each face of the cube is

$10 C$
$5 C$
$30 C$
$15 C$

Given cube encloses $\frac{30 C}{ϵ_{o}}$
we know gauss law $\oint E \cdot d a = \frac{q_{o}}{ϵ_{o}}$
$∴$ Total flux through all 6 faces $= 30 C$ per unit permittivity
$\Rightarrow$ Flux through each face $= \frac{30}{6}$

$\Rightarrow \frac{5 C}{ϵ_{o}}$