JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \(.........V cm\).
- A \(640\)
- B \(689\)
- C \(652\)
- D \(258\)
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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