Q.A closed triangular box is kept in an electric field of magnitude $E=2\times10^3\ \text{N C}^{-1}$, as shown in the figure (a vertical rectangular face of $15\ \text{cm}$ height, a slanted face, and a $5\ \text{cm}$ base, with the field making an angle of $60^\circ$ with the base). Calculate the electric flux through

Step 1. Flux through the vertical rectangular face (perpendicular to $E$, area $0.05\ \text{m}\times0.15\ \text{m}=0.0075\ \text{m}^2$): $\Phi_{\text{vertical}}=EA=2\times10^3\times0.0075=15\ \text{N m}^2\text{C}^{-1}$.

Step 2. The two triangular side faces of the box are parallel to $E$ (E lies in their plane), so they each carry zero flux.

Step 3. Since the closed box encloses no charge, the total flux through the whole surface must be exactly zero (Gauss's law); with the two triangular faces contributing nothing and the vertical face contributing $+15\ \text{N m}^2\text{C}^{-1}$ (outward), the slanted face must contribute $-15\ \text{N m}^2\text{C}^{-1}$ (inward) so the sum is zero -- equivalently, the slanted face ca …

Compute the vertical face's flux directly (EA), and use zero-net-flux for a charge-free closed …

• Forgetting the two triangular end faces are parallel to E and so carry zero flux -- only the vertic …