Question

there is uniform field 8×10^3iN/C. what is the net flux(in SI units) of the uniform electric field through a cube of side 0.3 m oriented so that its faces are parallel to the coordinate plane??

Answer 1

Answer:

The net flux will be $720 \dfrac{N-m^2}{C}$.

Explanation:

Given that,

Electric field $E=8\times10^{3}i N/C$

Side of cube $a=0.3\ m$

Area $A = 0.09\ m^2$

The flux is the dot product of electric field and area.

The net flux will be

$\phi=E\cdot s$

$\phi=(8\times10^{3}i +0j+0k)\cdot(0.09i+0.09j+0.09k)$

$\phi=8\times10^{3}\times0.09$

$\phi=720 \dfrac{N-m^2}{C}$

Hence, The net flux will be $720 \dfrac{N-m^2}{C}$.

Answer 2

Answer:0

Explanation:Mathematically electric flux is calculated by-

φ=∫E.dA

=E.A.cosθ

where, E- Electric field, A- Area vector of the cross section, θ- Angle between area vector and electric field.

Area vector of the faces of the cube are always taken perpendicular to their plane and in outward direction.

The electric field will enter and exit through the faces which are parallel to Y-Z axis.

Thus net flux can be calculated as,

= [3x10³(0.2x0.2)cos(180°)+3x10³(0.2x0.2)cos(0°)]

=(120-120)

=0

Thus the net flux through the cube will be zero as the flux due to both the parallel faces of the cube cancel each other because the area vector formed due to these faces are in opposite direction to each other.