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

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

Step-by-step explanation:

The expression for the electric flux through the cuboid is given by:

`\phi_E=E* A=\frac {q}{\epsilon_0}`

Since, area of cuboid = 6a²

`\phi_E=E* 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* a^2=\frac {q}{6* \epsilon_0}`

So,

Given ,

q = 5.0×10⁻¹² C

`\phi_E=\frac {5* 10^(-12)\ C}{6* 8.85* 10^(-12)\ C^2N^(-1)m^(-2)}`

Solving,

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