Question

A charge of 0.80 nC is placed at the center of a cube that measures 4.0 m along each edge. What is the electric flux through one face of the cube

Answer 1

The magnetic flux is
`\phi = 15 \ Nm^2 /C`

Step-by-step explanation:

From the question we are told that

The value of the charge is
`q = 0.80 \ nC = 0.80 *10^(-9) \ C`

The length of each side of the cube is
`d = 4.0 \ m`

Generally the magnetic flux through a closed surface is mathematically represented as

`\phi = (q)/(\epsilon_o) * D`

Where D is the area enclosing the charge

Now a cube is made up of six faces but in this question we are considering only one face which is mathematically represented as

`D = (1)/(6)`

So the electric flux through one face of the cube is mathematically represented as

`\phi = (q)/(6 * \epsilon _o )`

where
`\epsilon _o` is the permitivity of free space with value

`\epsilon_o = 8.85 *10^(-12) F/m`

substituting value

`\phi = (0.80 *10^(-9))/(6 * 8.85 *10^(-12) )`

`\phi = 15 \ Nm^2 /C`