The box-like Gaussian surface of Fig. 1 encloses a net charge of +24.0ε0 C and lies in an electric field given by N/C, with x and z in meters and b a constant. The bottom face is in the xz plane; the top face is in the horizontal plane passing through y2 = 1.00 m. For x1 = 1.00 m, x2 = 4.00 m, z1 = 1.00 m, and z2 = 3.00 m, what is b?
The electric field at one face of a parallelepiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field is also uniform over the entire face and is directed into that face (see Fig. 3). The two faces in question are inclined at 300 from the horizontal, while and are both horizontal, and have respective magnitudes of 2.50∙104 N/C and 7.00 ∙104 N/C. Assuming that no other electric field lines cross the surface of the parallelepiped, determine the net charge contained within.
We can't see your figures so I haven't a clue what 'b' is.
I also don't know that the å in
+24.0å0 is supposed to represent.
Obviously this is a Gauss' Law exercise. The flux of E out of the parallelipiped is proportional to the charge inside. There will be a å (permittivity) term in the equation, which is probably not the same meaning as yours.
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To find the value of b, we need to use Gauss's Law, which states that the electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space (ε0).
The first step is to calculate the electric flux through the Gaussian surface. The electric flux (Φ) is given by the integral of the dot product of the electric field (E) and the area vector (dA) over the surface:
Φ = ∫E · dA
Since the electric field is constant and parallel to the surface, the dot product reduces to a scalar multiplication:
Φ = E · A
Where A is the area of the Gaussian surface.
In this case, the area of the Gaussian surface is given by the product of the length (Δx = x2 - x1) and width (Δz = z2 - z1):
A = Δx * Δz
Therefore, the electric flux becomes:
Φ = E * (Δx * Δz)
We are given the electric field value (2.50 N/C), the Δx (4.00 m - 1.00 m = 3.00 m), and the Δz (3.00 m - 1.00 m = 2.00 m). Substituting the values:
Φ = 2.50 N/C * (3.00 m * 2.00 m)
Φ = 15 N * m²/C
Now, using Gauss's Law, we can equate the electric flux to the net charge enclosed divided by ε0:
Φ = Q_enclosed / ε0
Given that Q_enclosed = +24.0ε0 C, we can substitute the values:
15 N * m²/C = +24.0ε0 C / ε0
Now, let's cancel out the permittivity of free space term, ε0, on both sides:
15 N * m²/C = +24.0 C
Dividing both sides by 24.0 C:
15 N * m²/C / 24.0 C = b
Simplifying:
0.625 N * m²/C = b
Therefore, b is equal to 0.625 N * m²/C.