JEE Mains · Physics · STD 12 - 1. Electric charges and fields
An electric field \(\overrightarrow{\mathrm{E}}=4 \mathrm{x} \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\; \mathrm{N} / \mathrm{C}\) passes through the box shown in figure. The flux of the electric field through surfaces \(A B C D\) and \(BCGF\) are marked as \(\phi_{I}\) and \(\phi_{\mathrm{II}}\) respectively. The difference between \(\left(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}\right)\) is (in \(\left.\mathrm{Nm}^{2} / \mathrm{C}\right)\)

- A \(48\)
- B \(52\)
- C \(56\)
- D \(60\)
Answer & Solution
Correct Answer
(A) \(48\)
Step-by-step Solution
Detailed explanation
The flux passes through \(\mathrm{ABCD}(\mathrm{x}-\mathrm{y})\) plane is zero, because electric field parallel to surface. Flux of the electric field through surface \(BCGF\) \((y-z)\) At BCGF (electric field)…
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