Question

A point charge of 5.0 multiply.gif 10-12 C is located at the center of a cubical Gaussian surface. What is the electric flux through each face of the cube?

Answer 1

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:

$\phi_E=E\times A=\frac {q}{\epsilon_0}$

Since, area of cuboid = 6a²

$\phi_E=E\times 6a^2=\frac {q}{\epsilon_0}$

Where,

E is the electric field

a is the side of the cuboid

q is the charge

$\epsilon_0$ 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:

$\phi_E=E\times a^2=\frac {q}{6\times \epsilon_0}$

So,

Given ,

q = 5.0×10⁻¹² C

$\phi_E=\frac {5\times 10^{-12}\ C}{6\times 8.85\times 10^{-12}\ C^2N^{-1}m^{-2}}$

Solving,

The electric flux though each face of the cube is 0.0942 Nm²/C.