Question

A cubical region of the side $a$ has its center at the origin. It encloses three fixed point charges, $ - q{\text{ at }}\left( {0, - a/4,0} \right),{\text{ + 3q at }}\left( {0,0,0} \right)$ and $ - q{\text{ at }}\left( {0, + a/4,0} \right)$. Choose the correct option $\left( s \right)$.

$\left( a \right)$ The net electric flux crossing the plane $x = + a/2$ is equal to the net electric force crossing the plane $x = - a/2$.
$\left( b \right)$ The net electric flux crossing the plane $y = + a/2$ is more than the net electric force crossing the plane $y = - a/2$.
$\left( c \right)$ The net electric flux crossing the entire region is $\dfrac{q}{{{\varepsilon _0}}}$.
$\left( d \right)$ The net electric flux crossing the plane $z = + a/2$ is equal to the net electric flux crossing the plane $x = + a/2$.

Answer 1

Hint This question will be solved on the concept of gauss law. First of all, we will calculate the net flux and which will be calculated by the total charge upon the permittivity. And then we will see the flux through each of the axes.

Complete Step By Step Solution So first of all we will see the net electric flux through the entire region of the cube.
So, it will be equal to
$ \Rightarrow \dfrac{{ - q + 3q - q}}{{{\varepsilon _0}}}$
And on solving the above equation, we get
$ \Rightarrow \dfrac{q}{{{\varepsilon _0}}}$
Now we can see that the planes are symmetrically distributed about each of the planes.
So the flux passing through them will also be the same.
Therefore,
The net flux through $x = + a/2$and $x = - a/2$ will be equal
And the net flux through $y = + a/2$and $y = - a/2$ will also be the same.
Similarly, the net flux $z = \pm a/2$ will be equal.
Following above will be the net electric flux crossing through the plane.

Hence, the correct option for the following question will be the options $\left( a \right),\left( c \right),\left( d \right)$.

Additional information Gauss' law for electrostatics relates the electric motion through a shut surface to the net charge enclosed by the surface. This then is the “limitation” of Gauss's law, that it can be used to calculate the field only in special cases. We cannot use Gauss's law to calculate the field due to an electric dipole.

Note Gauss's law can be used to easily find the electric field due to a point charge, a spherically symmetric charge distribution, a uniform rod of charge, and a uniform sheet of charge. In each of these cases, we can draw a suitable Gaussian surface (or closed surface) which has the appropriate symmetry for the problem.