Electrostatics

JEE Main level, numerical answer, medium

270 of 311

Question 270 of 311

The electric field in a region is given by E→=(2i^+4j^+6k^)×103 N/C\overrightarrow{\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−zx-z plane is 6.0Nm2C−16.0 \mathrm{Nm}^2 \mathrm{C}^{-1}. The area of the surface is ________ cm2\mathrm{cm}^2.

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Answer: 15

ϕ=E→⋅A→=(2i^+4j^+6k^)×103⋅ Aj^6=4×103 A A=1.5×10−3 m2=15 cm2\begin{aligned} & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times 10^3 \cdot \mathrm{~A} \hat{\mathrm{j}} \\ & 6=4 \times 10^3 \mathrm{~A} \\ & \mathrm{~A}=1.5 \times 10^{-3} \mathrm{~m}^2 \\ & =15 \mathrm{~cm}^2\end{aligned}

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