Consider an electric field $\vec{E}=E_0 \hat{x}$, where $E_0$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is
$2 E_0 a^2$
$\sqrt{2} E_0 a^2$
$E_0 a^2$
$\frac{E_0 a^2}{\sqrt{2}}$
The figure shows a hollow hemisphere of radius $R$ in which two charges $3q$ and $5q$ are placed symmetrically about the centre $O$ on the planar surface. The electric flux over the curved surface is
Electric charge is uniformly distributed along a long straight wire of radius $1\, mm$. The charge per $cm$ length of the wire is $Q$ $coulomb$. Another cylindrical surface of radius $50$ $cm$ and length $1\,m$ symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is
A charge $Q$ is situated at the comer of a cube, the electric flux passed through all the six faces of the cube is
An electric field $\overrightarrow{\mathrm{E}}=4 \mathrm{x} \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\; \mathrm{N} / \mathrm{C}$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $BCGF$ are marked as $\phi_{I}$ and $\phi_{\mathrm{II}}$ respectively. The difference between $\left(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}\right)$ is (in $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$
A long cylindrical shell carries positive surface charge $\sigma$ in the upper half and negative surface charge $-\sigma$ in the lower half. The electric field lines around the cylinder will look like figure given in : (figures are schematic and not drawn to scale)