Practising Gauss law

elementary yet beautiful problem,

i dont think any one will need a solution for this

basically., let us cover the cylinder's ends with two spherical caps centred at origin,(charge)

the difference in the fluxes will reveal me the number of lines that have escaped through the surface of the cylinder

to find the flux we simply use the fact that it is a spherically symmetric charge and the fraction of flux through a cap is simply the fraction of area it covers times $\frac {q}{e}$ (where e is for "epsilon")

the values have been given such as to get all results in integers

the final answer is

$\frac {q(cos( \theta)- cos( \phi) )}{2e}$ where theta and phi are the subtended angles of the edges of cylinder with axis at the end points

which is 9/35 times total flux (use values)

@Karan Shekhawat - hope this clears it

The question is asking for the flux through the open cylinder defined by $x^2 +y^2 = 24$ , $1 \le z \le 5$ .

Consider a cylindrical element of length $dz$ at a distance $z$ along the axis of the given cylinder (from the origin). At all points on this element, the electric field is given by:

$\vec{E} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q}{z^2+r^2}$

However, this entire field does not contribute to the flux through the differential element; rather, it is the component of the field perpendicular to the surface that contributes to the flux:

$E \cos \theta = \dfrac{1}{4 \pi \epsilon_0} \dfrac{qr}{(z^2+r^2)^{\frac{3}{2}}}$

We know that the surface area of the cylindrical element is given by:

$dS = 2 \pi r dz$

Hence, the total electric flux through the given cylindrical surface is:

$\phi = \int_{1}^{5} \vec{E} \cdot \vec{dS}$

$\phi = \int_{1}^{5} E \cos \theta dS$

$\phi = \int_{1}^{5} \dfrac{2 \pi r dz}{4 \pi \epsilon_0} \dfrac{qr}{(z^2+r^2)^{\frac{3}{2}}}$

$\phi = \dfrac{qr^2}{2 \epsilon_0} \int_{1}^{5} \dfrac{dz}{(z^2+r^2)^{\frac{3}{2}}}$

$\Longrightarrow \phi = \boxed{\dfrac{9}{35} \dfrac{q}{\epsilon_0}}$