Q72 / 75 · JEE Main 2020 (09 Jan Shift 2) · Physics

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 $N m^{2} / C$ ) ____________.

Correct answer: 48

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}

Hence, the correct answer is 48.

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