What will be the total flux through the faces of the cube as in figure with side of length $'a'$ if a charge $'q'$ is placed at ?
$(a)$ $C$ $:$ centre of a face of the cube.
$(b)$ $D$ $:$ midpoint of $B$ and $C$.
What will be the total flux through the faces of the cube as in figure with side of length $'a'$ if a charge $'q'$ is placed at ?
$(a)$ $C$ $:$ centre of a face of the cube.
$(b)$ $D$ $:$ midpoint of $B$ and $C$.
$(a)$ If the charge $q$ is placed at $\mathrm{C}$, the centre of a face of the cube, it is being shared equally by 2 cubes.
$\therefore$ Flux through each cube,
$\phi=\frac{\phi^{\prime}}{\epsilon_{0}}=\frac{q}{2 \epsilon_{0}}$
$(b)$ Finally, if charge $q$ is placed at $\mathrm{D}$, the midpoint of $\mathrm{B}$ and $\mathrm{C}$, it is being shared equally by 2 cubes. $\therefore$ Flux through each cube,
$\phi=\frac{\phi^{\prime}}{\epsilon_{0}}=\frac{q}{2 \epsilon_{0}}$
Similar Questions
A circular disc of radius $R$ carries surface charge density $\sigma(r)=\sigma_0\left(1-\frac{r}{R}\right)$, where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is. . . . . .
A circular disc of radius $R$ carries surface charge density $\sigma(r)=\sigma_0\left(1-\frac{r}{R}\right)$, where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is. . . . . .
- [IIT 2020]
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:-
For a given surface the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this we can conclude that
For a given surface the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this we can conclude that
Draw electric field by negative charge.
Draw electric field by negative charge.
The electric field in a region of space is given by, $\overrightarrow E = {E_0}\hat i + 2{E_0}\hat j$ 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
The electric field in a region of space is given by, $\overrightarrow E = {E_0}\hat i + 2{E_0}\hat j$ 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
- [JEE MAIN 2014]