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Exercises · 1.17

Q.A point charge +10 μC+10\,\mu\text{C} is a distance 5 cm5\,\text{cm} directly above the centre of a square of side 10 cm10\,\text{cm}, as shown in Fig. 1.31. 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 cm10\,\text{cm}.)

A point charge 5 cm above the centre of a 10 cm square, used to find the electric flux through the square via Gauss's law
Figure 1.31
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Completing the square into a cube of edge 10 cm10\,\text{cm} places the charge at the cube's centre; Gauss's law gives total flux q/ε0q/\varepsilon_0, and by symmetry each of the six faces carries q/6ε0=1.88×105 N⋅m2/Cq/6\varepsilon_0=1.88\times10^{5}\,\text{N·m}^2/\text{C}.

A single square is an open surface, so Gauss's law cannot be applied to it directly. The hint tells us to complete it into a closed surface.

Step 1 — Build the cube. The charge sits 5 cm5\,\text{cm} above the centre of the 10 cm10\,\text{cm} square. Imagine a cube of edge 10 cm10\,\text{cm} having this square as one face. The centre of such a cube is 5 cm5\,\text{cm} from each face — exactly where the charge is. So the charge is at the centre of the cube, and the given square is one of its six faces.

Step 2 — Total flux through the cube. The cube is now a closed surface enclosing q=+10 μCq=+10\,\mu\text{C}. Gauss's law gives

Φtotal=qε0,ε0=8.854×10−12 C2/N⋅m2.\Phi_{\text{total}}=\frac{q}{\varepsilon_0},\qquad \varepsilon_0=8.854\times10^{-12}\,\text{C}^2/\text{N·m}^2.

Step 3 — Use symmetry. With the charge at the centre, the six faces are equivalent, so each receives one‑sixth of the total flux: …

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