A cubical volume is bounded by the surfaces x=0,x=a,y=0,y=a,z=0,z=a. The electric field in the region is given by E=E_0x i. Where E_0=4× 10^4 NC^-1…
JEE Main 2023 — Physics Electromagnetism
2023integerhard
A cubical volume is bounded by the surfaces x=0,x=a,y=0,y=a,z=0,z=a. The electric field in the region is given by E=E0xi^. Where E0=4×104NC−1m−1. If a=2cm, the charge contained in the cubical volume is Q×10–14C. The value of Q is ______.
(Take ϵ0=9×10−12C2N−1m−2)
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Held on 1 Feb 2023 · Verified 6 Jul 2026.
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Solution
Given here, electric field, E=E0xi^
Here, the flux passes mainly through surface areas, ABCDandEFGH. As the surfaces AEFBandCGHD are parallel to the electric field, so flux for these surfaces are zero. Again in EFGH, a=0, thus, electric field is zero.
Hence, the flux only passes through the surface are ABCD.
The net flux is ϕnet=ϕABCD=E0a⋅a2
Using Gauss's law, ϵ0qen=E0a3
⇒qen=E0ϵ0a3
=4×104×9×10−12×8×10−6
=288×10−14C
Hence, the value of Q=288.
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