GATE EC 2019 Set 1 — Question 21

The problem asks for the electric flux through the surface of a quarter-cylinder due to an infinite line charge along its axis.Let the linear charge density of the infinite line charge be $\lambda$. The problem states the charge density is Q. Based on the units in the options, Q must represent the linear charge density, so $\lambda = Q$ (in Coulombs per meter).According to Gauss's Law, the total electric flux $\Phi_{total}$ emanating from a charge $Q_{enc}$ enclosed within a closed surface is given by:$$\Phi_{total} = \oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$$Consider a Gaussian surface in the form of a full cylinder of height H and radius R, coaxial with the line charge. The charge enclosed by this full cylinder is $Q_{enc} = \lambda H = QH$.The total flux emanating from this segment of the line charge is:$$\Phi_{total} = \frac{QH}{\epsilon_0}$$Due to the cylindrical symmetry of the infinite line charge, the electric field $\vec{E}$ is directed radially outward from the line charge. That is, $\vec{E} = E(r) \hat{r}$.This total flux passes through the curved surface of the full cylinder. The flux through the top and bottom flat circular surfaces is zero because the electric field vector $\vec{E}$ is parallel to these surfaces (i.e., perpendicular to their area vectors).Now, consider the quarter-cylinder of height H. It subtends an angle of $\frac{\pi}{2}$ radians (or 90 degrees), which is $\frac{1}{4}$ of the full circle ($2\pi$ radians).By symmetry, the flux is distributed uniformly in the azimuthal direction. Therefore, the flux passing through the curved surface of the quarter-cylinder is exactly $\frac{1}{4}$ of the total flux.Flux through curved surface of quarter-cylinder = $\frac{1}{4} \Phi_{total} = \frac{1}{4} \left( \frac{QH}{\epsilon_0} \right) = \frac{QH}{4\epsilon_0}$.The question asks for the flux $\int \vec{E} \cdot d\vec{a}$ through the quarter-cylinder. This refers to the flux through the surface area of the quarter-cylinder. The surface consists of the curved part, two flat rectangular sides, and two flat quarter-circle ends (top and bottom).

• Flux through curved surface: As calculated above, it is $\frac{QH}{4\epsilon_0}$.
• Flux through flat rectangular sides: The area vectors of these sides are in the azimuthal direction ($\hat{\phi}$), while the electric field is in the radial direction ($\hat{r}$). Since $\hat{r} \cdot \hat{\phi} = 0$, the flux through these two sides is zero.
• Flux through top and bottom quarter-circle ends: The area vectors of these ends are in the axial direction ($\hat{z}$), while the electric field is in the radial direction ($\hat{r}$). Since $\hat{r} \cdot \hat{z} = 0$, the flux through these ends is also zero.

Therefore, the total flux through the entire surface of the quarter-cylinder is just the flux through its curved part, which is $\frac{QH}{4\epsilon_0}$.This matches option (B).