Electrostatics

JEE Main level, numerical answer, medium

299 of 311

Question 299 of 311

An electric field E⃗=4xi^−(y2+1)j^N/C\vec{E}=4x\hat{i}-\left({y}^{2}+1\right)\hat{j}N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCDABCD and BCGFBCGF are marked as ϕI{\phi }_{\text{I}} and ϕII{\phi }_{\text{II}} respectively. The difference between (ϕI−ϕII)\left({\phi }_{I}-{\phi }_{II}\right) is (in Nm2/CN{m}^{2}/C ) ____________.

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Answer: 48

Flux via ABCDABCD
ϕ1=∫E.dA=0{\phi }_{1}=\int E.dA=0
Flux via BCEFBCEF
ϕ2=∫E.dA{\phi }_{2}=\int E.dA
ϕ2=E.A=(4xi^−(y2+1)j^).4i^{\phi }_{2}=E.A=\left(4x\hat{i}-\left({y}^{2}+1\right)\hat{j}\right).4\hat{i}
=16xwhere,x=3=16x where,x=3
ϕ2=48N.m2Cϕ1−ϕ2=−48N.m2C{\phi }_{2}=48\dfrac{N.{m}^{2}}{C} {\phi }_{1}-{\phi }_{2}=-48\dfrac{N.{m}^{2}}{C}

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