JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2020
- Medium
- MCQ
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 \(\left.\mathrm{Nm}^{2} / \mathrm{C}\right)\)
Pick an option to check your answer.
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) \(\Rightarrow \overrightarrow{\mathrm{E}}=12 \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\)
\((x=3 m)\)
Flux \(\phi_{\mathrm{II}}=12 \times 4=48 \mathrm{Nm}^{2}{ / \mathrm{C}}\)
So \(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}=0-48=-48 \mathrm{Nm}^{2}{ / \mathrm{C}}\)
\(\therefore\) Correct answer \(-48\)
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