Question

A particle of charge $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\mu C}$is at the center of a Gaussian cube$\mathbf{55}\mathbf{}\mathbf{cm}$on edge. What is the net electric flux through the surface?

Short answer

The net electric flux through the surface is $2.0\times {10}^5\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

Step-by-step solution

The given data

1. Charge of the particle,$\mathrm{q}=1.8\mathrm{\mu C}$
2. Edge length of cube,$\mathrm{a}=55\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.55\mathrm{m}$

Understanding the concept of Gauss law-planar symmetry

Using the concept of Gauss law and the planar symmetry, we can get the required values of the electric field at the left of the plates, right of the plates, and between the plates.

Formula:

The total flux through any surface, $\mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (1)

Calculation of the net flux through any cube surface

As the cube has six surfaces, thus, the net flux through each surface of the cube is given using equation (1) such that,

$$\begin{gathered} \mathrm{\varphi }=\frac{1.8\times {10}^{-6}\mathrm{C}}{6\times \left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)} \\ =2.0\times {10}^5\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the required flux is$2.0\times {10}^5\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$