Question

A charged shell of radius R carries a total charge Q. Given $\phi$ as the flux of electric field through a closed cylindrical surface of height h, radius r and with its centre same as that of the shell. Here the centre of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following are correct?
A) if h>2R, r>R, then $\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$
B) if $h<\dfrac{8R}{5},r=\dfrac{3R}{5},\phi =0$
C) $h>2R,r=\dfrac{4R}{5},\phi =\dfrac{Q}{5{{\varepsilon }_{0}}}$
D) $h>2R,r=\dfrac{3R}{5},\phi =\dfrac{Q}{5{{\varepsilon }_{0}}}$

Answer 1

Let us find the electric flux for all given cases in the above question. Gauss law can be used to find out the electric flux flowing through the surface. We need to apply all the conditions given in the options and find the electric flux using the gauss law.

Complete step by step answer:
Let us write down the given values in option a and solve for electric flux using gauss law.
$\begin{aligned} & given \\\ & h>2R,r>R \\\ \end{aligned}$
From gauss law,
$\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$
In option b,
$\begin{aligned} & h=\dfrac{8R}{5},r=\dfrac{3R}{5} \\\ & \phi =0 \\\ \end{aligned}$
The cylinder made using the parameters will exactly fit into the circle. The electric flux inside the circle is zero. So, the flux is equal to zero in option b.
In option c,
$h=2R,r=\dfrac{4R}{5}$
If we look at the picture given below,
The charge enclosed in the shaded region will be,
$\begin{aligned} & {{q}_{enc}}=2\pi \times \dfrac{1-\cos 53\times Q}{4\pi } \\\ & \Rightarrow {{q}_{enc}}=\dfrac{2Q}{5} \\\ & \therefore \phi =\dfrac{2Q}{5{{\varepsilon }_{0}}} \\\ \end{aligned}$
In option d,
$h=2R,r=\dfrac{3R}{5}$
Similar to option c, the charge can be calculated as,
$\begin{aligned} & {{q}_{enc}}=2\times 2\pi (1-\cos 37)\dfrac{Q}{4\pi } \\\ & \Rightarrow {{q}_{enc}}=\dfrac{Q}{5} \\\ & \therefore \phi =\dfrac{Q}{5{{\varepsilon }_{0}}} \\\ \end{aligned}$
Hence, the correct options are option a,b and d.

Additional information:
Electric flux is the rate of flow of the electric field through a given area (see ). Electric flux is proportional to the number of electric field lines going through a virtual surface. Electric flux has SI units of volt metres (V m). Gauss’s law is a law relating the distribution of electric charge to the resulting electric field. Gauss’s law is one of the four Maxwell’s equations which form the basis of classical electrodynamics. Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. An electric field is a region of space in space around a charged particle or in between two voltages. It exerts force on charged objects in its vicinity.

Note:
In the above question, we need to note that the electric flux is zero outside the surface where there is no electric field. Simply, the electric flux is not affected by charges that are not within the closed surface. In the case of an electric field, it will get affected by charges that are outside the closed surface too.