A cubical volume is bounded by the surfaces x = 0 ,   x = a ,   y = 0 ,   y = a ,   z =…

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=4×104 NC-1 m-1. If a=2 cm, the charge contained in the cubical volume is Q×1014 C. The value of Q is ______.

(Take ϵ0=9×10-12 C2 N-1m-2)

Solution

Given here, electric field, E=E0xi^

Here, the flux passes mainly through surface areas, ABCD and EFGH. As the surfaces AEFB and CGHD are parallel to the electric field, so flux for these surfaces are zero. Again in EFGHa=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, qenϵ0=E0a3

qen=E0ε0a3

=4×104×9×10-12×8×10-6

=288×10-14 C

Hence, the value of Q=288.

Asked in: JEE Main 2023 (01 Feb Shift 2)

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