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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2023
  • Medium
  • Numerical

Expression for an electric field is given by \(\vec{E}=4000 x^2 \hat{i} \frac{V}{m}\). The electric flux through the cube of side \(20\,cm\) when placed in electric field (as shown in the figure) is \(.........V cm\).

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Answer & Solution

Correct Answer

(A) \(640\)

Step-by-step Solution

Detailed explanation

Flux \(=\overrightarrow{ E } \cdot \overrightarrow{ A }\)

\(=4000(0 \cdot 2)^2 \frac{ V }{ m } \cdot(0 \cdot 2)^2 m ^2\)

\(=4000 \times 16 \times 10^{-4}\,Vm\)

\(=640\,V\,cm\)

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