The figure shows a cube whose sides are all 5cm long. Over the entire left-hand side of the cube is a uniform electric field E1=7500 N/C which is entering the cube at 20 degrees below the x axis. Over the right-handed side is a uniform electric field E2=4500 N/C which is leaving the cube at 60 degrees below the x axis.
a) Assuming that no other electric field lines cross the surface of the cube, determine the net charge contained within.
b) Is the electric field produced only by the charge within the cube, or is the field also due to charges outside? How can you tell?
a. Gauss' law. E dot Area=chargeenclosed
b. Is there a component of E on each side the same? If so, that E is generated outside.
For instance, on the left side, E dot area=7500*25cos20=1.76E5
one the right side, leaving 4500*25cos60=1.07E5
So the meaning of this is that there is a net flux entering, which means a charge is enclosed, and the remainder of E going to the right is caused by something exterior to the cube.
Is the magnitude the same as the net charge for a?
yes, the net charge comes from E dot A
To determine the net charge contained within the cube, we need to use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is directly proportional to the charge enclosed by that surface.
a) To apply Gauss's Law, we need to select a closed surface whose electric flux can be easily calculated. In this case, we can choose the entire cube as the closed surface.
Since the electric field lines enter the left-hand side of the cube and exit the right-hand side, we can infer that the electric field lines on the top and bottom faces cancel each other out, resulting in no net electric flux through these faces.
Therefore, the only contribution to the electric flux comes from the front face, which has an area of (5 cm) × (5 cm) = 25 cm^2.
The electric flux through the front face of the cube can be calculated using the formula:
Electric flux = Electric field × Area × cos(angle between field and normal vectors)
Given that the electric field on the front face (E1) is 7500 N/C and the angle between the electric field vector and the normal vector of the front face is 20 degrees below the x-axis, we can calculate the electric flux:
Electric flux = 7500 N/C × 25 cm^2 × cos(20 degrees)
b) To determine whether the electric field produced is only due to charges within the cube or if there are charges outside, we need to analyze Gauss's Law further.
According to Gauss's Law, the net electric flux through the closed surface is equal to the sum of the electric fluxes due to charges within the surface and charges outside the surface.
If the net flux through the closed surface is non-zero, it implies the presence of charges outside the surface contributing to the electric field. On the other hand, if the net flux is zero, it suggests that the electric field is solely due to charges within the surface.
In this case, after calculating the electric flux through the front face of the cube, if the value is non-zero, it indicates the electric field is not solely due to charges within the cube. Conversely, if the value is zero, it suggests that the electric field is solely due to charges within the cube.
By comparing the calculated electric flux with zero, we can determine whether the electric field produced is solely due to charges within the cube or if there are external charges contributing to it.