A closed Gaussian surface in the shape of a cube of edge length 1.9 m with one corner at x = 1 4.8 m, y = 1 3.9 m.
The cube lies in a region where the electric field vector is given by NC with y in meters.
What is the net charge contained by the cube?

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

50 inches should be the resultant vector

the correct choice is a .

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