Question

A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 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 cm.)

Answer 1

$\Phi \approx 1.88 \times 10^5\, \text{N m}^2/\text{C}$

Answer 2

Solution Explanation:
Construct a cube with the given square as one face (edge = 10 cm). Then the cube’s center lies 5 cm below that face. Since the charge +10 μC is located 5 cm directly above the centre of the square, it coincides with the center of the cube. By symmetry, the total electric flux from a point charge is Q/ε₀ and is equally distributed through all 6 faces. Hence, the flux through one face (i.e. the given square) is:

$$\Phi = \frac{Q}{6 \varepsilon_0}$$

Substituting the values:

• $Q = 10 \times 10^{-6}\, \text{C}$
• $\varepsilon_0 = 8.85 \times 10^{-12}\, \text{C}^2/(\text{N}\cdot\text{m}^2)$

$$\Phi = \frac{10 \times 10^{-6}}{6 \times 8.85 \times 10^{-12}} \approx 1.88 \times 10^5\, \text{N m}^2/\text{C}$$