Question

A point charge $+10\,\mu\text{C}$ is a distance $5\,\text{cm}$ directly above the centre of a square of side $10\,\text{cm}$, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10\,\text{cm}$.)

Hint:

Since 5 cm is half of the 10 cm side, treat the square as one face of a 10 cm cube with the charge at its centre. By symmetry the flux through one face is (1/6)(q/epsilon_0).

Answer 1

A point charge \(q = +10\,\mu\text{C} = 1\times10^{-5}\,\text{C}\) sits \(5\,\text{cm}\) directly above the centre of a square of side \(10\,\text{cm}\).

Step 1: Set up the cube trick. The charge is at height \(5\,\text{cm}\), which is exactly half of the square's side (\(10\,\text{cm}\)). Imagine a cube of edge \(10\,\text{cm}\) with the charge at its centre; the given square is then one of the six faces of this cube.

Step 2: Total flux through the cube (Gauss's law).

\[\phi_{total} = \frac{q}{\varepsilon_0}, \qquad \varepsilon_0 = 8.85\times10^{-12}.\]

Step 3: Symmetry. The charge sits at the centre, so by symmetry the flux is shared equally among all six faces. The flux through one face (our square) is

\[\phi = \frac{1}{6}\,\frac{q}{\varepsilon_0}.\]

Step 4: Substitute and compute.

\[\phi = \frac{1}{6} \times \frac{1\times10^{-5}}{8.85\times10^{-12}}\]\[\phi = \frac{1}{6} \times 1.13\times10^{6}\]\[\phi = 1.88\times10^{5}\,\text{N m}^{2}/\text{C}.\]

The result is independent of the square's side because the charge being at the cube centre is what fixes the one-sixth share.

\[\boxed{\phi \approx 1.88\times10^{5}\,\text{N m}^{2}/\text{C}}\]

Answer 2

Solid-angle approach:

Step 1: Flux from a point charge through a surface depends on the solid angle \(\Omega\) that the surface subtends at the charge:

\[\phi = \frac{q}{4\pi\varepsilon_0}\,\Omega = \frac{q}{\varepsilon_0}\cdot\frac{\Omega}{4\pi}.\]

Step 2: A full sphere around the charge subtends the total solid angle \(4\pi\) steradians. When the charge is at the centre of a cube, each of the six identical faces subtends an equal slice of that total, namely

\[\Omega_{face} = \frac{4\pi}{6} = \frac{2\pi}{3}.\]

Step 3: Therefore the flux through one face is

\[\phi = \frac{q}{\varepsilon_0}\cdot\frac{1}{6}.\]

Step 4: Putting in numbers with \(q = 1\times10^{-5}\,\text{C}\) and \(\varepsilon_0 = 8.85\times10^{-12}\):

\[\phi = \frac{1\times10^{-5}}{6 \times 8.85\times10^{-12}} = 1.88\times10^{5}\,\text{N m}^{2}/\text{C}.\]

The square caught one-sixth of all the field lines leaving the charge.

\[\boxed{\phi \approx 1.88\times10^{5}\,\text{N m}^{2}/\text{C}}\]