Problem 32 - A Complete Resource Book in Physics for JEE Main

Question

A cylinder of radius \(R\) and length \(L\) is placed in a uniform electric field \(E\) parallel to the axis of cylinder. The total flux through the curved surface of the cylinder is given by (A) \(2 \pi R^{2} E\) (B) \(2 \pi R^{2} / E\) (C) \(E 2 \pi R L\) (D) Zero

Step-by-Step Solution

Answer

The total electric flux through the curved surface of the cylinder is zero because the electric field is parallel to the axis of the cylinder. As a result, the angle between the electric field vector and the area vector of each infinitesimal area on the surface is \(90^\circ\), and their dot product is zero. Integrating the electric flux over the entire surface gives a total of zero. So, the correct answer is (D) Zero.

Step 1: Understanding Gauss's Law and application

Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\(\varepsilon_0\)). Mathematically, it can be written as: \[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}\] In this problem, we have a cylinder with no charge enclosed within it, so the charge enclosed (\(Q_{enclosed}\)) is zero. In that case, Gauss's Law simplifies to: \[\oint \vec{E} \cdot d\vec{A} = 0\] We will apply this law on the curved surface of the cylinder.

Step 2: Dividing the curved surface into infinitesimal areas

To apply Gauss's law, we need to divide the curved surface of the cylinder into infinitesimal areas. Consider an infinitesimal area \(dA\) on the curved surface of the cylinder. The vector \(d\vec{A}\) representing this area is perpendicular to the surface and aligned along the electric field line. Since the electric field is parallel to the axis of the cylinder, we can see that the angle between the electric field vector \(\vec{E}\) and the area vector \(d\vec{A}\) will be \(90^\circ\).

Step 3: Calculating the electric flux through the infinitesimal area

Since the angle between the electric field vector and the area vector is \(90^\circ\), the dot product of the electric field vector and the area vector is: \[\vec{E} \cdot d\vec{A} = E \cdot dA \cdot \cos{90^\circ} = E \cdot dA \cdot 0 = 0\] Thus, the electric flux through the infinitesimal area \(dA\) is zero.

Step 4: Integrating over the curved surface of the cylinder

Now, we need to integrate the electric flux over the entire curved surface of the cylinder to find the total electric flux. Since the electric flux through each infinitesimal area is zero, the integral will also be zero: \[\oint \vec{E} \cdot d\vec{A} = 0\] This means that the total electric flux through the curved surface of the cylinder is zero. So, the correct answer is: (D) Zero