JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 2)
An electric field $$\overrightarrow E = 4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$$ N/C
passes through the box shown in figure. The
flux of the electric field through surfaces ABCD
and BCGF are marked as $${\phi _I}$$ and $${\phi _{II}}$$
respectively. The difference between $$\left( {{\phi _I} - {\phi _{II}}} \right)$$ is (in Nm2/C) _______._9th_January_Evening_Slot_en_2_1.png)
passes through the box shown in figure. The
flux of the electric field through surfaces ABCD
and BCGF are marked as $${\phi _I}$$ and $${\phi _{II}}$$
respectively. The difference between $$\left( {{\phi _I} - {\phi _{II}}} \right)$$ is (in Nm2/C) _______.
Answer
-48
Explanation
Flux via ABCD
$$\phi $$I = $$\int {\overrightarrow E } .d\overrightarrow A $$ = 0
Flux via EFGH
$$\phi $$II = $$\int {\overrightarrow E } .d\overrightarrow A $$
= [$$4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$$].4$$\widehat i$$
= 16x = 16 $$ \times $$ 3 = 48
$${\phi _I} - {\phi _{II}}$$ = 0 - 48 = -48 Nm2/C
$$\phi $$I = $$\int {\overrightarrow E } .d\overrightarrow A $$ = 0
Flux via EFGH
$$\phi $$II = $$\int {\overrightarrow E } .d\overrightarrow A $$
= [$$4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$$].4$$\widehat i$$
= 16x = 16 $$ \times $$ 3 = 48
$${\phi _I} - {\phi _{II}}$$ = 0 - 48 = -48 Nm2/C
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