Question

Electric field in a region is given by $E=-4x\hat i+6y\hat j$, find the charge enclosed in the cube side \[1m\] oriented as shown in figure.

\[\begin{align}
  & A.2{{\varepsilon }_{0}} \\
 & B.zero \\
 & C.{{\varepsilon }_{0}} \\
 & D.6{{\varepsilon }_{0}} \\
\end{align}\]

Answer 1

Hint: Here we have a cube, which is clearly a closed figure. Electric field is the electric force due to a unit positive charge which is at rest would exert on its surrounding. We need to calculate the charge in the cube. But we know that the total charge enclosed in a closed surface is given by Gauss law.

Formula used:
\[{{\Phi }_{_{E}}}=\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\nolimits_E
 E\cdot dA \]

Complete answer:
We know that the electric force due to a pair of charges is given by Coulomb's law. An electric field can be produced by a time-varying electric field or an electrical charge. These can be either attracting or repelling in nature.
An electric field $E$ is defined as the electric force $F$ per unit positive charge$q$, which is infinitesimally small and at rest, and is given as
$E=\dfrac{F}{q}$
From Gauss law, we know that the total electric flux through a closed surface is equal to $\dfrac{1}{\epsilon_{0}}$ times the charge enclosed in the surface, and it is given by
$\Phi_{E}=\dfrac{q}{\epsilon_{0}}$
Gauss law can also be given as \[{{\Phi }_{_{E}}}=\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\nolimits_E
 E\cdot dA \] where \[dA\] is the surface area vector. This form of Gauss law is called the integral form of Gauss law.
Given that $E=-4x\hat i+6y\hat j$ and the charge enclosed in cube side\[1m\], implies $x=y=z=1$
Then $\Phi=6ydy-4xdx=6\times 1\times 1=4\times 1\times 1=6-4=2$
Then from $\Phi_{E}=\dfrac{q}{\epsilon_{0}}$ we get,$2=\dfrac{q}{\epsilon_{0}}$
Or,$q=2\epsilon_{0}$

Hence the answer is A.$2\epsilon_{0}$.

Note:
Electric field is in the direction of the force. Usually, the electric field of a point positive charge is radially outwards, whereas the electric field of a point negative charge is radially inwards to the charge. However, the electric field also depends on the symmetry of the charge carrying conductor.