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

    $6.30$

  • B

    $6.35$

  • C

    $6.40$

  • D

    $6.45$

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  • [JEE MAIN 2022]

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 ?

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An electric field is uniform, and in the positive $x$ direction for positive $x,$ and uniform with the same magnitude but in the negative $x$ direction for negative $x$. It is given that $E =200 \hat{ i }\; N/C$ for $x\,>\,0$ and $E =  - 200\hat i\;N/C$ for $x < 0 .$ A right ctrcular cyllnder of length $20 \;cm$ and radius $5\; cm$ has its centre at the origin and its axis along the $x$ -axis so that one face is at $x=+10\; cm$ and the other is at $x=-10\; cm$

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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