Question:
A point charge of $+12 \mu \mathrm{C}$ is at a distance $6 \mathrm{~cm}$ vertically above the centre of a square of side $12 \mathrm{~cm}$ as shown in figure. The magnitude of the electric flux through the square will be ____________$\times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}$.
Solution:
From symmetry $\phi=\frac{1}{6}\left(\frac{\mathrm{q}}{\varepsilon_{0}}\right)$
$=\frac{12 \times 10^{-6}}{6 \times 8.85 \times 10^{-12}}$
$=225.98 \times 10^{3} \frac{\mathrm{Nm}^{2}}{\mathrm{~s}}$
$\simeq 226 \times 10^{3} \frac{\mathrm{Nm}^{2}}{\mathrm{C}}$