Answer:
The electric flux though each face of the cube is 0.0942 Nm²/C.
Explanation:
The expression for the electric flux through the cuboid is given by:
[tex]\phi_E=E\times A=\frac {q}{\epsilon_0}[/tex]
Since, area of cuboid = 6a²
[tex]\phi_E=E\times 6a^2=\frac {q}{\epsilon_0}[/tex]
Where,
E is the electric field
a is the side of the cuboid
q is the charge
[tex]\epsilon_0[/tex] is the constant having value 8.85×10⁻¹² C²/Nm²
Thus, the expression for the electric flux through one face of the cuboid is given by:
[tex]\phi_E=E\times a^2=\frac {q}{6\times \epsilon_0}[/tex]
So,
Given ,
q = 5.0×10⁻¹² C
[tex]\phi_E=\frac {5\times 10^{-12}\ C}{6\times 8.85\times 10^{-12}\ C^2N^{-1}m^{-2}}[/tex]
Solving,
The electric flux though each face of the cube is 0.0942 Nm²/C.
