Question

A cube encloses 24 coulomb .The electric flux passing through each surface of the cube is.....

Answer 1

Answer:

24/6ε0. Here ε0=8.85*10^-12

Explanation:

Gauss's Law. The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.

flux=q/6ε0

=>24/6ε0

Answer 2

Given:

Charged enclosed by cube $\[ = 24\,C\]$

To Find: Electric flux passing through each surface of the cube

Solution:

The gauss law states that if a surface encloses $q$ charge, then the net flux passing through the surface is $\phi = \frac{q}{{{\varepsilon _0}}}$.

As the cube encloses the charge $\[24\,C\]$, therefore, the net flux coming out from the cube is given as:

$\begin{gathered} \phi = \frac{{24C}}{{{\varepsilon _0}}} \hfill \\ \phi = \frac{{24}}{{8.85 \times {{10}^{ - 12}}}} \hfill \\ \phi = 2.71 \times {10^{12}}N{m^2}{C^ - }^1 \hfill \\ \end{gathered}$

Since the cube has six faces with equal area, therefore, the flux coming out from each surface can be given as:

$\begin{gathered} {\phi _{each\,\,surface}} = \frac{\phi }{6} \hfill \\ {\phi _{each\,\,surface}} = \frac{{2.71 \times {{10}^{12}}}}{6} \hfill \\ {\phi _{each\,\,surface}} = 4.51 \times {10^{11}}N{m^2}{C^ - }^1 \hfill \\ \end{gathered}$

Hence, the electric flux passing through each surface of the cube is $4.51 \times {10^{11}}N{m^2}{C^ - }^1$.

#SPJ2