CBSE Class 12-science Answered
The electric field components in are Ex = A+Bx, Ey = Ez = 0, in which A = 800 N/C , B= 100 N/C-m. Calculate
(a) the flux through the cube
(b) the charge within the cube. Assume that a = 0.1 m
Asked by hitanshu04 | 22 May, 2021, 07:41: PM
Expert Answer
Let us assume that the cube is positioned in Cartesian coordinate system so that centre of the cube
is at the origin and sides are parallel to the axes.
Hence the face ABCD is at x = +a/2 and the face EFGH is at x = -a/2as shown in figure.
Flux φ passing through the surface of cube is determined as
where 
is electric field vector , dA is area element and integration is performed over the closed surface of cube
is electric field vector , dA is area element and integration is performed over the closed surface of cubeSince electric field is only in x-axis direction , Flux is determined as
............................... (1)where E1x is electric field at face ABCD , i.e., at x = +a/2 , E2x is electric field at face EFGH , i.e., at x = -a/2
First term in eqn.(1) is positive because unit normal vecrtor is in +x axis direction .
Second term in eqn.(1) is negative because unit normal vecrtor is in -x axis direction .
E1x = ( A + 0.05 × B )
E2x = ( A - 0.05 × B )
Hence eqn.(1) is written as ,
( surface integration will give the face area of the cube )
By Gauss law , enclosed charge Q is related to the surface flux φ as , φ = Q / εo
where εo is permittivity of free space
Enclosed charge Q = ( φ × εo ) = ( 0.1 × 8.854 × 10-12 ) C = 8.854 × 10-11 C
Answered by Thiyagarajan K | 23 May, 2021, 12:47: AM
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