In finding the electric field using Gauss Law the formula $|\overrightarrow{\mathrm{E}}|=\frac{q_{\mathrm{enc}}}{\varepsilon_{0}|\mathrm{A}|}$ is applicable. In the formula $\varepsilon_{0}$ is permittivity of free space, $A$ is the area of Gaussian surface and $q_{enc}$ is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?
Only when the Gaussian surface is an equipotential surface
Only when $|\overrightarrow{\mathrm{E}}|=$ constant on the surface.
For any choice of Gaussian surface.
Only when the Gaussian surface is an equipotential surface and $|\overrightarrow{\mathrm{E}}|$ is constant on the surface.
Draw electric field lines when two positive charges are near.
A charge $+q$ is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius $R_1$ and outer radius $R_2$. A charge $+Q$ is placed at a distance $r > R_2$ from the centre of the shell. Then the electric field in the hollow cavity
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
How does the no. of electric field lines passing through unit area depend on distance ?
A cubical volume is bounded by the surfaces $x =0, x = a , y =0, y = a , z =0, z = a$. The electric field in the region is given by $\overrightarrow{ E }= E _0 \times \hat{ i }$. Where $E _0=4 \times 10^4 NC ^{-1} m ^{-1}$. If $a =2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$. The value of $Q$ is $...........$
Take $\left.\varepsilon_0=9 \times 10^{-12} C ^2 / Nm ^2\right)$