A point charge of 1.8 uC is at the center of a Gaussian cube 55 cm on edge. What is the net electric flux through the surface?
I have no idea where to start with this problem.
According to Gauss' Law, you will get the same answer no matter where the charge q is placed inside the box, and no matter how large the box is. It does not even have to be a cube, as long as it encloses the charge.
Total flux = q/(epsilon-zero)
= 1.8*10^-6/8.82*10^-12
= 2.04*10^5 (N*m^2/C)
That is a lot easier than doing the actual integhtration of E(normal) over the six sides of the cube, but the answer will be the same.
Can u pls explain with examples
To solve this problem, you can use Gauss's Law, which states that the net electric flux through a closed surface is equal to the charge enclosed divided by the electric constant.
Here are the steps to solve this problem:
Step 1: Find the charge enclosed within the Gaussian cube. Since the point charge of 1.8 μC is at the center of the cube, it is fully enclosed within the cube itself. Therefore, the charge enclosed is 1.8 μC.
Step 2: Calculate the electric constant. The electric constant, also known as the permittivity of free space, is denoted by ε₀ and has a value of approximately 8.85 x 10⁻¹² C²/N m².
Step 3: Calculate the area of each face of the cube. Since the cube is symmetrical, each face has the same area. The area of a face of the cube is equal to the square of the length of one edge. Therefore, the area of each face is (0.55 m)² = 0.3025 m².
Step 4: Calculate the total area of all six faces of the cube. Since there are six faces, you can find the total area by multiplying the area of one face by six. Therefore, the total area of all faces is 6 x 0.3025 m² = 1.815 m².
Step 5: Calculate the net electric flux through the surface using Gauss's Law. The net electric flux (Φ) is equal to the charge enclosed (Q) divided by the electric constant (ε₀), multiplied by the total area of the surface (A). Mathematically, it can be written as Φ = (Q / ε₀) * A.
Plugging in the values, you have:
Φ = (1.8 x 10⁻⁶ C / (8.85 x 10⁻¹² C²/N m²)) * 1.815 m².
Calculating this expression will give you the net electric flux through the surface of the Gaussian cube.
To find the net electric flux through the surface of the Gaussian cube, you can use Gauss's Law, which states that the electric flux through a closed surface is proportional to the net charge enclosed by that surface.
The formula to calculate the electric flux through a closed surface is:
Φ = Electric Field * Surface Area * cos(θ)
However, since the cube is symmetrical and the charge is at the center, we only need to consider one face of the cube. We can then multiply the electric flux through one face by the total number of faces to get the net electric flux.
To calculate the electric field at the surface of the cube, we can use the formula for the electric field due to a point charge:
E = k * (q / r^2)
Where:
- E is the electric field
- k is the electrostatic constant (8.99 x 10^9 Nm²/C²)
- q is the charge (1.8 uC = 1.8 x 10^-6 C)
- r is the distance from the charge to the surface (half the length of the edge of the cube, which is 55 cm = 0.55 m)
Plugging in these values, we can calculate the electric field at the surface of the cube.
Once we have the electric field, we can calculate the electric flux through one face of the cube using the formula mentioned earlier:
Φ = E * Surface Area * cos(θ)
Where:
- Surface Area is the area of one face of the cube (side length squared)
- cos(θ) is the angle between the electric field and the normal to the surface (which is 0 degrees in this case as the electric field is perpendicular to the surface)
Finally, multiply the electric flux through one face by the total number of faces (6) to obtain the net electric flux.