CBSE 2025, Set 3

Question

An electric field $\vec{E}$ given by :
$$\vec{E} = 100\hat{i}\frac{	ext{N}}{	ext{C}} \quad 	ext{for } x > 0$$
$$= -100\hat{i}\frac{	ext{N}}{	ext{C}} \quad 	ext{for } x < 0$$
exists in a region. A right circular cylinder of length $10	ext{ cm}$ and radius $2	ext{ cm}$, is placed in the region such that its axis coincides with x-axis and its two faces are at $x = -5	ext{ cm}$ and $x = 5	ext{ cm}$. Calculate:
(a) the net outward flux through the cylinder, and
(b) the net charge inside the cylinder.

Edvaya · Accepted Answer

The correct answer is Subjective. **(a) Net outward flux through the cylinder:**
The electric field points outward through both flat faces of the cylinder. The curved surface is parallel to the electric field, so the flux through it is zero.

Flux through the left face ($x = -5	ext{ cm}$):
$$\phi_L = \vec{E} \cdot \Delta\vec{S} = 100(-\hat{i}) \cdot \Delta S(-\hat{i})$$
$$\phi_L = 100 	imes \pi (2 	imes 10^{-2})^2$$
$$\phi_L = 4\pi 	imes 10^{-2}	ext{ Nm}^2/	ext{C}$$

Flux through the right face ($x = +5	ext{ cm}$):
$$\phi_R = \vec{E} \cdot \Delta\vec{S} = 100(\hat{i}) \cdot \Delta S(\hat{i})$$
$$\phi_R = 100 	imes \pi (2 	imes 10^{-2})^2$$
$$\phi_R = 4\pi 	imes 10^{-2}	ext{ Nm}^2/	ext{C}$$

Total outward flux:
$$\phi_{total} = \phi_L + \phi_R = 4\pi 	imes 10^{-2} + 4\pi 	imes 10^{-2}$$
$$\phi_{total} = 8\pi 	imes 10^{-2}	ext{ Nm}^2/	ext{C}$$
$$\phi_{total} = 25.12 	imes 10^{-2}	ext{ Nm}^2/	ext{C}$$

**(b) Net charge inside the cylinder:**
Using Gauss's Law, $\phi_{total} = \frac{q}{\epsilon_0}$
$$q = \epsilon_0 \phi_{total}$$
$$q = (8.854 	imes 10^{-12}) 	imes (25.12 	imes 10^{-2})$$
$$\boxed{q = 0.22 	imes 10^{-11}	ext{ C}}$$

> The flux through the curved surface is zero because the area vector is perpendicular to the electric field.

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