Find the electric field due to an infinite plane of positive charges with uniform surface charge density sigma
E = sigma / 2 eo
why:
take field along axis at distance x from uniformly charged disk radius R,
at radius r
dQ = 2 pi sigma r dr
so dEx = (1/4 pi eo) [ (2 pi sigma r dr)x ] / (x^2+r^2)^1.5
integrate from 0 to R
Ex = (sigma x/ 2eo) integral r dr/ (x^2+r^2)^1.5
= (sigma /2 eo) [ 1 - 1/ sqrt( {R^2/x^2}+1) ]
if R >>x then
Ex = sigma / 2eo
Ah, an infinite plane of positive charges? How electrifying! Well, let me enlighten you with a humorous explanation.
Imagine this infinite plane as a gigantic dance floor, filled with positively charged partygoers. Now, the uniform surface charge density sigma can be seen as the number of charges per unit area.
Just like at a party, these charges will create an electric field around them. It's like they're all trying to attract you to the dance floor! The electric field created by this plane will be parallel to the surface and always pointing away from it.
To determine the electric field strength, we can use a mathematical trick called Gauss's law. It tells us that the electric field created by a large, flat group of charges is the charge density divided by two times the permittivity of space, which is approximately 8.854 x 10^-12 C^2/(N*m^2).
So, the electric field (E) can be calculated as E = sigma / (2 * ε₀). But remember, my dear friend, that the direction of the electric field will always be perpendicular to the plane.
That's the shocking truth about the electric field due to an infinite plane of positive charges! Enjoy the electrifying dance floor!
To find the electric field due to an infinite plane of positive charges with a uniform surface charge density sigma, you can use Gauss's Law.
Step 1: Choose a Gaussian surface parallel to the plane of charges. This surface should be a cylinder with one circular face inside the region containing the charges and the other face outside the region.
Step 2: Calculate the electric flux through the Gaussian surface. Since the electric field is constant and parallel to the surface, the electric flux passing through the curved surface of the cylinder is zero. The only contribution comes from the flat circular face inside the region containing the charges.
Step 3: Apply Gauss's Law, which states that the electric flux passing through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε0). The charge enclosed by the Gaussian surface is equal to the surface charge density (sigma) multiplied by the area of the circular face.
Step 4: Equate the electric flux passing through the circular face of the Gaussian surface to the expression from Gauss's Law to find the electric field:
Electric flux = Electric field * Area = (sigma * A) / ε0
Where A is the area of the circular face.
Step 5: Solve for the electric field:
Electric field = (sigma / ε0)
So the electric field due to an infinite plane of positive charges with uniform surface charge density sigma is equal to sigma divided by the permittivity of free space (ε0).
To find the electric field due to an infinite plane of positive charges with a uniform surface charge density sigma, you can use Gauss's law.
Step 1: Choose a Gaussian surface that is symmetric with respect to the plane of charges. A good choice is a cylindrical Gaussian surface with its axis perpendicular to the plane of charges.
Step 2: Apply Gauss's law, which states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε0). In this case, the closed surface is the cylindrical Gaussian surface.
Step 3: Calculate the total charge enclosed by the Gaussian surface. Since the surface charge density sigma is uniform, the charge enclosed is equal to sigma multiplied by the area of the Gaussian surface.
Step 4: Calculate the electric field by using the equation: Electric field (E) multiplied by the area of the curved surface of the Gaussian surface equals the charge enclosed divided by ε0. Since the electric field is constant and perpendicular to the curved surface, the electric field is simply E multiplied by the area of the curved surface.
Step 5: Solve for the electric field (E) by rearranging the equation from step 4: E = sigma / (2ε0)
Therefore, the electric field due to the infinite plane of positive charges with uniform surface charge density sigma is E = sigma / (2ε0).