Question

In Figure, a proton is a distance d/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.)

Answer 1

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 $$q=+1.6\times10^{-19}\, C $$ situated at the inside center of the cube. The cube has six faces, and we expect an equal amount of flux through each face. The total amount of flux is Φnet = q/ε0, and we conclude that the flux through the square is one-sixth of that. Thus,

$$\phi=\dfrac{q}{6\epsilon_0}=\dfrac{1.6\times10^{-19}\, C}{6(8.85 \times10^{12} \,C^2/N.m^2 )}=3.01\times10^{-9}\,N.m^2/C$$