Question 305: Electrostats & Capacitance

Question

An infinitely long wire has uniform linear charge density $\lambda = 2 	ext{ nC/m}$. The net flux through a Gaussian cube of side length $\sqrt{3}$ cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be $x 	ext{ Nm}^2	ext{C}^{-1}$, where $x$ is:
[Neglect any edge effects and use $\frac{1}{4\pi \epsilon_0} = 9 	imes 10^9$ SI units]

Accepted Answer

To find the net flux through a Gaussian cube when an infinitely long wire with a uniform linear charge density $ \lambda = 2 	ext{ nC/m} $ passes through the cube's diagonal corners, we use Gauss's Law. The relevant expression is:
$ \phi = \frac{q_{	ext{enc}}}{\varepsilon_0} $
Here, $ q_{	ext{enc}} $ is the charge enclosed by the Gaussian surface, and $ \varepsilon_0 $ is the permittivity of free space.
Given:
The wire passes through two opposite corners of the cube.
The side length of the cube, $ a = \sqrt{3} 	ext{ cm} = \sqrt{3} 	imes 10^{-2} 	ext{ m} $.
The length of the wire inside the cube is equal to its diagonal, which is $ \sqrt{3} 	imes a $.
Thus, the charge enclosed $ q_{	ext{enc}} $ is:
$ q_{	ext{enc}} = \lambda 	imes 	ext{(length of wire inside the cube)} = \lambda 	imes \sqrt{3} 	imes a $
Substitute the given values:
$ \lambda = 2 	imes 10^{-9} 	ext{ C/m}, \quad a = \sqrt{3} 	imes 10^{-2} 	ext{ m} $
Calculating:
$ q_{	ext{enc}} = 2 	imes 10^{-9} 	imes \sqrt{3} 	imes \sqrt{3} 	imes 10^{-2} $
Now apply Gauss's Law for net flux:
$ \phi = \frac{\lambda \cdot \sqrt{3} \cdot a}{\varepsilon_0} $
Substitute $\varepsilon_0 = \frac{1}{4\pi 	imes 9 	imes 10^9}$:
$ \phi = 2 	imes 10^{-9} 	imes \sqrt{3} 	imes \sqrt{3} 	imes 10^{-2} 	imes 36 \pi 	imes 10^9 $
$ \phi = 2.16 \pi 	ext{ Nm}^2	ext{C}^{-1} $
Thus, the net flux through the Gaussian cube is $ 2.16 \pi 	ext{ Nm}^2	ext{C}^{-1} $.

Answer

$6.48 \pi$

Answer

$0.72 \pi$

Answer

$1.44 \pi$

Answer

$2.16 \pi$

Electrostats & Capacitance

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