Final answer:
To find the charge contained inside the parallelepiped, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. By choosing a Gaussian surface that encloses one face of the parallelepiped and calculating the electric flux on that face, we can find the charge enclosed by the surface.
Step-by-step explanation:
In order to find the charge contained inside the parallelepiped, we can use Gauss's law.
Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space.
In this case, since the parallelepiped is not closed, we can choose a Gaussian surface that encloses only one face of the parallelepiped. This face has an area of l1 * l2.
The electric field on this face is Eleft to the left of the dashed line and Eright to the right of the dashed line. The flux through this face is then equal to Eleft * l1 * l2 to the left of the dashed line and Eright * l1 * l2 to the right of the dashed line.
Since the electric field is constant on this face, we can take the average of Eleft and Eright to find the flux. The average electric field is (Eleft + Eright) / 2.
Using Gauss's law, we have (Eleft + Eright) / 2 * l1 * l2 = qenclosed / ε0, where qenclosed is the charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
Solving for qenclosed, we get qenclosed = (Eleft + Eright) * l1 * l2 * ε0 / 2.
Plugging in the values of Eleft = -55.37 N/C, Eright = 55.37 N/C, l1 = 23.82 cm = 0.2382 m, l2 = 11.27 cm = 0.1127 m, and the value of ε0, which is approximately 8.85 × 10^-12 N^-1 m^-2 C^2, we can calculate the charge contained inside the parallelepiped.