If $\vec E = \frac{E_0 x}{a} \hat i$,then find the electric flux through the shaded area of the cube as shown in the figure,where the shaded face is at $x = a$.

The electric field in a region of space is given by $\overrightarrow E = E_0 \hat i + 2E_0 \hat j$,where $E_0 = 100 \, N/C$. The flux of the field through a circular surface of radius $0.02 \, m$ parallel to the $Y-Z$ plane is nearly:

What is the net flux of the uniform electric field of $E = 3 \times 10^{3} \hat{i} \; N/C$ through a cube of side $20 \; cm$ oriented so that its faces are parallel to the coordinate planes?

An electric field is given by $\vec{E} = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \ N/C$. The electric flux through a surface area $\vec{A} = 30 \hat{i} \ m^2$ (in $SI$ units) is:

An electric field $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j} \text{ N/C}$ passes through the box shown in the figure. The flux of the electric field through surfaces $ABCD$ and $BCGF$ are marked as $\phi_I$ and $\phi_{II}$ respectively. The difference $(\phi_I - \phi_{II})$ is (in $\text{Nm}^2/C$):

For a given surface,the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this,we can conclude that: