Question

A charge Q is located inside a rectangular box. Theelectric flux through each of the six surfaces of the box is: electric flux 1 = +1500 N*m^2/C, electric flux 2 = +2200N*m^2/C, electric flux 3 = +4600 N*m^2/C, electric flux 4 = -1800N*m^2/C, electric flux 5 = -3500 N*m^2/C, and electric flux 6 =-5400 N*m^2/C. What is Q?

Answer 1

To solve this problem we will apply the laws of gaus that relate the Electric Flow as the charge of the object on the permittivity constant of free space. Mathematically this is

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

Rearranging to find the charge,

`Q = \phi \epsilon_0`

Here

Q = Charge

`\phi =` Electric Flux

`\epsilon_0 =` Permittivity of free space

The total flux would be

`\phi_T = \phi_1+\phi_2+\phi_3+\phi_4+...+\phi_(\infty)`

`\phi = ( 1500+2200+4600-1800-3500-5400 ) N\cdot m^2 / C`

`\phi = - 2400 N\cdot m^2 / C`

Replacing we have that,

`Q = (-2400 N\cdot m^2/C)( 8.85*10^(-12) C^2 / N \cdot m^2)`

`Q = -21240 * 10^(-12) C`

`Q = - 21.24 nC`

Therefore the charge Q inside a rectangular box is -21.24nC