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PhysicsElectrostaticsGauss LawEasy2 minPYQ_2020

PhysicsEasynumerical

An electric field $\vec{E} = 4 x \hat{i} - (y^{2} + 1) \hat{j} N / C$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $B C G F$ are marked as $ϕ_{I}$ and $ϕ_{II}$ respectively. The difference between $(ϕ_{I} - ϕ_{I I})$ is (in $N m^{2} / C$ ) ____________.

Answer:

48.00

Solution:

Flux via $A B C D$
$ϕ_{1} = \int E . d A = 0$
Flux via $B C E F$
$ϕ_{2} = \int E . d A$
$ϕ_{2} = E . A = (4 x \hat{i} - (y^{2} + 1) \hat{j}) . 4 \hat{i}$
$= 16 x w h e r e, x = 3$
$ϕ_{2} = 48 \frac{N . m^{2}}{C} ϕ_{1} - ϕ_{2} = - 48 \frac{N . m^{2}}{C}$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Gauss Law

⏱ 2mℹ️ Source: PYQ_2020

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