Question

A point charge 8.85 µC is at a distance 25 cm directly above the centre of a square of side length 50 cm. What is the magnitude of electric flux through the square?

• A. 1.67 x 104 V m
• B. 1.67 x 105 V m
• C. 1.67 x 106 V m
• D. 106 V m

Answer 1

Answer: (B)

1.67 x 105 V m

Answer 2

We can construct an imaginary cube of side length 50 cm having the given square as one of its faces. The charge is 25 cm above the square, which is exactly half the edge length, so the charge is at the center of the cube. By symmetry, the total electric flux from the point charge is

$$\Phi_{\text{total}} = \frac{Q}{\varepsilon_0}$$

and is equally divided among the 6 faces, so the flux through one face is

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

Substitute the values:

$$Q = 8.85 \times 10^{-6}\,\text{C}, \quad \varepsilon_0 = 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)$$ $$\Phi = \frac{8.85 \times 10^{-6}}{6 \times 8.85 \times 10^{-12}} = \frac{1}{6}\times10^6 \approx 1.67 \times 10^5\,\text{N m}^2/\text{C}.$$

Since 1 N m²/C is equivalent to 1 V m, the answer is $1.67 \times 10^5$ V m.