A linear charge rho L = 2.0 uC lies on the y-z plane. Find the electric flux passing through the plane extending from 0 to 1.0m in the x direction and from -infinity to infinity in the y direction.
Alright, so the formula to use is:
flux = (integral(integral)(Ds * dS)) where dS is a vector whose direction is normal to the surface element ds and has a magnitude of ds and Ds is the electric flux density passing through ds.
Ds is usually = Q / (4pi*R^2)
I have no idea what to make dS or how to start this. Any hints/tips?
To find the electric flux passing through the given plane, you can use Gauss's Law. Gauss's Law relates the electric flux passing through a closed surface to the total charge enclosed by that surface.
In this case, the charge is a linear charge distribution, given by rho L = 2.0 uC. To apply Gauss's Law, we need to choose a closed surface that encloses the given linear charge. In this problem, the most convenient choice is a rectangular box with one side parallel to the x-axis, extending from 0 to 1.0m, and the other two sides extending infinitely in the y and z directions.
Now let's go step by step to solve this problem:
Step 1: Choose a Gaussian surface:
In this case, we choose a rectangular box with one side parallel to the x-axis, extending from 0 to 1.0m, and the other two sides extending infinitely in the y and z directions.
Step 2: Determine the charge enclosed:
To find the charge enclosed by the Gaussian surface, we need to calculate the total charge for the given linear charge distribution. Since the linear charge distribution is along the y-z plane, the charge enclosed is the charge density (rho L = 2.0 uC) multiplied by the length of the line segment within the Gaussian surface (which is 1m). So the charge enclosed is simply 2.0 uC.
Step 3: Determine the electric flux:
The electric flux passing through the Gaussian surface is given by the formula:
flux = (integral(integral)(Ds * dS))
Where dS is a vector whose direction is normal to the surface element ds and has a magnitude of ds, and Ds is the electric flux density passing through ds.
In this case, the electric flux density Ds is given by Ds = Q / (4pi*R^2), where Q is the charge enclosed by the Gaussian surface and R is the distance from the charge to the surface element ds.
Step 4: Calculate the electric flux:
To calculate the electric flux passing through the Gaussian surface, we integrate the electric flux density (Ds) over the surface elements (dS). Since the box is symmetric and the electric field due to a linear charge decreases with the square of the distance, the electric flux passing through the top and bottom surfaces of the box will cancel out. Therefore, we only need to calculate the flux passing through the side surfaces (left and right sides of the box).
Since the charge is uniform along the y-z plane, we can assume that the magnitude of the electric field is the same at all points on the side surfaces. Taking the magnitude of the electric field to be E, the electric flux through each side surface can then be calculated as E * area, where the area is the product of the length of the side surface and the height (which extends infinitely in the y-direction).
Step 5: Simplify and solve:
Calculate the electric flux passing through the side surface of the box by following these steps:
- Determine the area of one side surface of the box: the length of the box is given as 1.0m, and the height extends infinitely in the y-direction.
- Calculate the magnitude of the electric field at the side surface points using the formula Ds = Q / (4pi*R^2), where R is the distance from the charge to the surface element ds.
- Multiply the magnitude of the electric field by the area of the side surface to obtain the electric flux for one side surface.
- Multiply the electric flux for one side surface by 2 to take into account both sides of the box.
Finally, add up the electric flux passing through the side surfaces of the box. The result will give you the electric flux passing through the plane extending from 0 to 1.0m in the x-direction and from -infinity to infinity in the y-direction.