Question

The electric field components in the following figure are Ex = αx, Ey = 0, Ez= 0; in which α = 400 N/C m. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a = 0.1m.

Answer 1

The electric flux is ϕNet = 1/∈0 (qin)  and the charge within the cube is  3.540 x 10-12 C.

Explanation:

Ex = ∝ x = 400 x  Ey = Ez = 0  

Hence flux will exist only on left and right faces of cube as Ex ≠ 0

EL = a^2(n2) + ER .a^2(nR) = 1/εo (qin) =  Φ

-EL .a^2(n2) +a^2 ER = ΦNet

ΦNet = - (400 a)a^2 + a^2 (400 x 2a)

= - 400 a^3 + 800 a^3  

= 400 a^3  

= 400 × (.1)3

ϕNet = 0.4 N.m^2c-1

ϕNet = 1/∈0 (qin)

qin = ∈0ϕNet

= 8.85 x 10-12 x 0.4 = 3.540 x 10-12 C

Thus the electric flux is ϕNet = 1/∈0 (qin)  and the charge within the cube is  3.540 x 10-12 C.

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