(b) In the example presented in the video, the electric flux was calculated along the sides of an imaginary rectangular box drawn between two charged plates, as shown in the image bele was true about the electric flux calculated on the top and bottom surfaces of the box? + + + + + + + + + + + + The electric flux on the top surface and bottom surface had opposite signs and noticeably different magnitudes. The electric flux on the top surface and bottom surface were equal in both magnitude and sign. The electric flux on the bottom surface had the same magnitude but opposite sign as the flux on the top surface. O The electric flux on the top surface and bottom surface were both equal to zero. (c) In the same example from the video described above, what was true about the electric flux calculated on the left and right surfaces of the box? The electric flux on the left surface and right surface were both equal to zero. The electric flux on the left surface and right surface were equal in both magnitude and sign. Both were nonzero. The electric flux on the left surface and right surface had opposite signs and noticeably different magnitudes. The electric flux on the left surface had the same magnitude but opposite sign as the flux on the right surface. Both were nonzero. (d) Which of the following statements best describes the relationship between the electric flux and electric charge for the example of the rectangular box presented in the video? The electric flux was zero on the left and right sides of the box, and nonzero on the top and bottom sides. The total electric flux was therefore nonzero, and so there must have been some electric charge inside the box. The electric field on all sides of the box was zero, so the total electric flux was zero, and there was no charge inside the box. Electric field was measured on the surfaces of the box. Therefore, there must have been some electric charge inside the box. Even though there was electric field measured on the surfaces of the box, the total electric flux was zero, and there was no charge inside the box. (e) Which statement below best describes the usefulness of Gauss's law? OGauss's law can never be used to find the electric field. It only can be used to find the electric flux. Gauss's law can always be used to find the electric field, no matter the charge distribution. Gauss's law can sometimes be used to find the electric field, provided the Gaussian surface is rectangular. Gauss's law can sometimes be used to find the electric field, particularly in cases where the field is zero, constant, or highly symmetric.