Net charge = 2.59nC
Explanation:
Gauss' Law states that the net electric flux is given by:
∮→E⋅→d/A = q enc/ϵ0
At this point, we have to solve the net electric flux through each side.
One corner is at (4.8, 3.9, 0).
The other corners are each 1.9m apart, so the other corners are at
(4.8, 5.8, 0), (6.7, 3.9, 0) and (6.7, 5.8, 0).
The other four corners are 1.9m away in the z-axis:
(4.8, 5.8, 2), (6.7, 3.9, 2) and (6.7, 5.8, 2).
Therefore, there are two planes (parallel to the x-z plane), one at
y = 3.9
and the other at
y = 5.8 which have a constant y-coordinate and are facing in the −^j and +^j respectively.
Area = 1.9m * 1.9m = 3.61m²
The flux through the first plane (area of 3.61m²) is given by:
E.A = (3.4i + 4.4 * 3.9²j + 3.0k) * (-3.61m²j) = -241.59564
The flux through the other plane is
E.A = (3.4i + 4.4 * 5.8²j + 3.0k)* (3.61m²j) = 534.33776
Now, for the other planes. There are no ^j components for the unit vectors for the area.
Therefore, even though they change in y-coordinate, those terms cancel out.
Therefore, for the planes with unit vector in the x-direction, we get:
E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±12.274
And in the z-direction:
E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±10.83
Now, when we sum all these fluxes together, the contribution from the x- and z-directions cancel out. Therefore, our net flux is:
-241.59564 + 534.33776 = 292.74212
Therefore, the enclosed charge is given by
q enc = ϵ0* (292.74212)
= 2.5856871136363E−9C
= 2.59E-9 nC--_- Approximated
= 2.59nC
