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An electric field \hat{E}=4\; x\hat{i}-(y^{2}+1)\hat{j}\; N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as  FI  and FII respectively. The difference between (FI-FII) is (in Nm^{2}/C)______.
Option: 1 48
Option: 2  38
Option: 3 58
Option: 4 62
 

Answers (1)

best_answer

 

 

 

 Flux via ABCD 

   \phi _{1}=\int \vec{E}\cdot d\vec{A}=0

Flux via BCEF 

\phi _{2}=\int \bar{E}\cdot d\bar{A}

\phi _{2}= \bar{E}\cdot \bar{A}

=(4x\hat{i}-(y^{2}+1)\hat{j})\cdot 4\hat{i}

=16x,x=3

\phi_{2}=48 \frac{N-m^{2}}{C}

\left | \phi _{1}-\phi _{2} \right |=48\frac{N-m^{2}}{C}

 

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