The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is
The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is
- A
$\phi = \frac{1}{{2{ \in _0}}}$
- B
$\phi = \frac{{ - 1}}{{2{ \in _0}}}$
- C
$\phi = \frac{{ - 1}}{{{ \in _0}}}$
- D
$\phi = \frac{1}{{{ \in _0}}}$
Similar Questions
A positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is flux through closed gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is flux through closed gaussian surface $S_2$ whose radius is just greater than $R_1$ then:-
A positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is flux through closed gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is flux through closed gaussian surface $S_2$ whose radius is just greater than $R_1$ then:-
Gauss's law can help in easy calculation of electric field due to
Gauss's law can help in easy calculation of electric field due to
A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$ -axis are shown in the figure. These lines suggest that:-
A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$ -axis are shown in the figure. These lines suggest that:-
Write $SI$ unit of electric flux.
Write $SI$ unit of electric flux.
Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface the electric field will be due to
Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface the electric field will be due to
- [IIT 2004]