Question

In Fig. 23-33, a proton is a distance $\mathit{d}\mathbf{/}\mathbf{2}$directly above the center of a square of side d. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge d.)

Short answer

The magnitude of the electric flux through the square is$3.01\times {10}^{-9}\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

Step-by-step solution

The given data

1. The Square of the side is$d$
2. Proton is placed at $\frac{\mathrm{d}}{2}$distance above the square

Understanding the concept of Gauss law-planar symmetry

The net flux through each cube surface is given by dividing the total flux value through the cube by 6. Thus, the net flux is given using the concept of the Gauss flux theorem.

Formula:

The net flux passing through an enclosed volume, ${\mathrm{\varphi }}_{\mathrm{net}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (1)

Calculation of the net flux through a square surface

To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the shape of a cube, of edge length d, with a proton of charge$\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$situated at the inside center of the cube.

The cube has six faces, and we expect an equal amount of flux through each face. Thus, the flux through the square is one-sixth of that the total flux of equation (1) and is given by:

$\begin{array}{l}\mathrm{\varphi }=\frac{1.6\times {10}^{-19}\mathrm{C}}{6\times (8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2)}\\ =3.01\times {10}^{-9}\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}\end{array}$

Hence, the value of the flux is $3.01\times {10}^{-9}\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.