Question

A cylinder of radius $R$ and the length $L$ is placed in the uniform electric field $E$ parallel to the cylinder axis. The total flux of the surface of the cylinder is given by,
a. $2\pi {R^2}E$
b. $\dfrac{{\pi {R^2}}}{E}$
c. $\dfrac{R}{E}$
d. Zero

Answer 1

We can solve the given problem with the help of the electric flux. We can calculate the electric flux at both the surfaces and then we add those values of flux to get the answer.

Formula used:
$\Rightarrow \phi = \int {\vec E.} d\vec s$
Where $\phi$is the electric flux and $E$ is the electric field.

Complete step by step answer:
In the question it is given that a cylinder of radius $R$ and the length $L$ is placed in the uniform electric field $E$ parallel to the cylinder axis.
We can represent these values in the diagram and we can take the diagram as a reference.
Consider the given diagram. In the diagram the green coloured line is the Electric field and it is parallel to the axis of the given cylinder. We have the radius and it is denoted by the blue colour. We have the surface area of the cylinder on the outside surface.
We can solve the given problem with the help of the electric flux. The electric flux can be defined as, the total number of electric field lines that passes through the given area by unit time.
$\Rightarrow \phi = \int {\vec E.} d\vec s$
Where $\phi$is the electric flux and $E$ is the electric field.
We can calculate the electric flux at both the surfaces and then we add those values of flux to get the answer.
The flux of the flat surface 1 is,
$\Rightarrow {\phi _1} = \int {\vec E.} d\vec s$
The value for the area and we can substitute in the equation. we get,
$\Rightarrow {\phi _1} = - E\left( {\pi {R^2}} \right)$
We have a minus sign because the area and the electric field is opposite.
The flux of the flat surface 2 is,
$\Rightarrow {\phi _2} = \int {\vec E.} d\vec s$
The value for the area and we can substitute in the equation. we get,
$\Rightarrow {\phi _2} = E\left( {\pi {R^2}} \right)$
The net flux the surface of the cylinder is given by,
$\Rightarrow \phi = {\phi _1} + {\phi _2}$
$\Rightarrow \phi = - E\left( {\pi {R^2}} \right) + E\left( {\pi {R^2}} \right)$
We get zero because the values are the same but the sign is different. The answer is,
$\therefore \phi = 0$

Hence, the correct answer is option (D).

Note: Remember that the unit of the electric flux voltmeters ($Vm$) or $N{m^2}{c^{ - 1}}$. The dimensional analysis is $M{L^3}{T^3}{A^{ - 1}}$.
Another method of calculating is:
Consider $\phi_1=E\times\pi{R^2}$
$\phi_2=-E\times\pi{R^2}$
Then the flux through the curved surface will be,
$C=\int Eds$
$\Rightarrow C=\int Eds \text{Cos}{90^\circ}=0$
$\therefore$ Total flux through cylinder $= \phi_1+\phi_2=0$