Question

Charges of 3.0 nC, -2.0 nC, -7.0 nC, and 1.0 nC are contained inside a rectangular box with length 1.0 m, width 2.0 m, and height 2.5 m. Outside the box are charges of 1.0 nC and 4.0 nC. What is the electric flux through the surface of the box?

Answer 1

To solve this problem we will apply Gauss's law which defines the electric flow as the proportional change of the charge on the vacuum permittivity. Mathematically said this is,

`\phi = (q)/(\epsilon_0)`

Here,

q = Charge

`\epsilon_0`= Vacuum permittivity

We will start calculating the load inside the box

`q_(enclosed) = \text{Charge inside the box}`

`q_(enclosed) = (3-2-7+1)nC`

`q_(enclosed) = -5nC`

Now if the vacuum permittivity is equivalent to,

`\epsilon_0 = 8.85*10^(-12) F/m`

We can replace in our first equation:

`\phi = (-5*10^(-9))/(8.85*10^(-12))`

`\phi = 565N\cdot m`

Therefore the electric flux through the surface of the box is
`565N\cdot m`