The electric field in a region is given by vecmathrmE = (2hatmathrmi + 4hatmathrmj + 6hatmathrmk) times 10^3mathrmN / mathrmC$\vec{\mathrm{E}} = (2\hat{\mathrm{i}} + 4\hat{\mathrm{j}} + 6\hat{\mathrm{k}}) \times 10^{3}\mathrm{N} / mathrm{C}$ . The flux of the field through a rectangular surface parallel to x-z plane is 6.0mathrmNm^2mathrmC^-1$6.0\mathrm{Nm}^2\mathrm{C}^{-1}$ . The area of the surface is __________ mathrmcm^2$\mathrm{cm}^2$ . [cite: 195, 196]

### Related Formula phi = vecE cdot vecA$$\phi = \vec{E} \cdot \vec{A}$$ [cite: 827] ### Core Logic A surface aligned parallel to the xtext-z$x\text{-}z$ plane possesses an area vector pointing completely orthogonal to it along the y$y$-axis direction, meaning vecA = Ahatj$\vec{A} = A\hat{j}$[cite: 196, 827]. Performing the dot product: [cite: 827] phi = left[(2hati + 4hatj + 6hatk) times 10^3right] cdot (Ahatj) = 4 times 10^3 A$$\phi = \left[(2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3\right] \cdot (A\hat{j}) = 4 \times 10^3 A$$ [cite: 195, 827] Given that the net flux magnitude is 6.0\ textNm^2textC^-1$6.0\ \text{Nm}^2\text{C}^{-1}$ [cite: 196]: 6 = 4 times 10^3 A implies A = frac64 times 10^3 = 1.5 times 10^-3\ textm^2$$6 = 4 \times 10^3 A \implies A = \frac{6}{4 \times 10^3} = 1.5 \times 10^{-3}\ \text{m}^2$$ [cite: 828, 829] Converting square meters to square centimeters (1\ textm^2 = 10^4\ textcm^2$1\ \text{m}^2 = 10^4\ \text{cm}^2$): [cite: 196, 830] A = 1.5 times 10^-3 times 10^4 = 15\ textcm^2$$A = 1.5 \times 10^{-3} \times 10^4 = 15\ \text{cm}^2$$ [cite: 830] ### Pattern Recognition Always focus exclusively on the specific field component matched to the surface orientation normal[cite: 827]. For an xtext-z$x\text{-}z$ plane match, only the hatj$\hat{j}$ coefficient creates flux[cite: 196, 827]. Do not miss the metric scale unit transition at the end (m^2 rightarrow cm^2$m^2 \rightarrow cm^2$)[cite: 196, 830]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics