Question

A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three negative charges, each has a charge q 2 2 = − 2.60 × 10 − 12 2.60×10−12 C as the drawing shows. What is the electric flux passing through the surface?(The permittivity of free space ε 0 0 = 8.85×10-12C²/(N.m²))

Answer 1

The electric flux passing through the given cubical Gaussian surface is zero.

Step-by-step explanation:

The electric flux passing through the given cubical Gaussian surface can be calculated using Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε0). In this case, we have two positive charges (+q1) and three negative charges (-q2) enclosed by the surface.

To calculate the total charge enclosed, we need to find the net charge by summing up the charges. Since the positive charges cancel out the negative charges, the net charge enclosed is zero (qenc = 0). Therefore, the electric flux passing through the surface is also zero.

So, the electric flux passing through the given cubical Gaussian surface is zero.

Answer 2

The electric flux is zero because charge is zero.

Step-by-step explanation:

Given that,

Positive charge
`q_(1)=3.90*10^(-12)\ C`

Negative charge
`q_(2)=-2.60*10^(-12)\ C`

We need to calculate the total charged

Using formula of charge

`Q_(enc)=2q_(1)+3q_(2)`

Put the value into the formula

`Q_(enc)=2*3.90*10^(-12)+3*(-2.60*10^(-12))`

`Q_(enc)=0`

We need to calculate the electric flux

Using formula of electric flux

`\phi=(Q_(enc))/(\epsilon_(0))`

Put the value into the formula

`\phi=(0)/(8.85*10^(-12))`

Hence, The electric flux is zero because charge is zero.