An electric field E → = 4 x i ^ - y 2 + 1 j ^ N / C passes through the box shown in figure. The flux…

An electric field E=4xi^-y2+1j^N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI and ϕII respectively. The difference between ϕI-ϕII is (in Nm2/C ) ____________.

Solution

Flux via ABCD
ϕ1=E.dA=0
Flux via BCEF
ϕ2=E.dA
ϕ2=E.A=4xi^-y2+1j^.4i^
=16x where, x=3
ϕ2=48N.m2C ϕ1-ϕ2=-48N.m2C

Asked in: JEE Main 2020 (09 Jan Shift 2)

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