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Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 6m.
To find the electric flux density (D) at (R = 6m) for a configuration involving multiple charged cylindrical sheets, we employ the concept of Gauss's law. Gauss's law in the context of electric flux density states that the net electric flux through a closed surface is equal to the charge enclosed byRead more
To find the electric flux density (D) at (R = 6m) for a configuration involving multiple charged cylindrical sheets, we employ the concept of Gauss’s law. Gauss’s law in the context of electric flux density states that the net electric flux through a closed surface is equal to the charge enclosed by that surface.
Given:
1. (sigma = 5 , text{C/m}^2) at (R = 2m)
2. (sigma = -2 , text{C/m}^2) at (R = 4m)
3. (sigma = -3 , text{C/m}^2) at (R = 5m)
And we are asked to find the flux density (D) at (R = 6m).
To start, we calculate the charge per unit length (lambda) on each cylindrical sheet using the surface charge density (sigma) and the relation (lambda = sigma cdot 2pi R) because the charge per unit length for a cylindrical shell of radius (R) is the product of the surface charge density and the circumference of the cylinder.
1. For (R = 2m): (lambda_1 = 5 cdot 2pi (2) = 20pi , text{C/m})
See less2. For (R = 4m):
Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m
To find the electric flux density ((D)) at (R = 4.5) m in the presence of three charged cylindrical sheets, we need to consider the charge densities ((sigma)) and their respective positions. The given cylindrical sheets have charge densities (sigma = 5), (sigma = -2), and (sigma = -3) C/m² at radiiRead more
To find the electric flux density ((D)) at (R = 4.5) m in the presence of three charged cylindrical sheets, we need to consider the charge densities ((sigma)) and their respective positions. The given cylindrical sheets have charge densities (sigma = 5), (sigma = -2), and (sigma = -3) C/m² at radii (R = 2) m, (R = 4) m, and (R = 5) m, respectively.
In cylindrical coordinates, given a uniformly charged infinite line or cylindrical sheet, the electric field ((E)) outside the cylinder, by Gauss’s law, relates to the enclosed charge. However, for non-conducting sheets, we deal directly with the surface charge and determine the electric displacement (flux density, (D)), which in free space and for linear materials is directly proportional to the electric field, with (D = epsilon_0 E), where (epsilon_0) is the vacuum permittivity ((8.854 times 10^{-12}) C²/N·m²).
However, because we are dealing with charges on cylindrical sheets and not points or infinite lines and considering the superposition principle (if there are multiple charges, the total electric field is the vector sum of the fields created by each charge alone), the approach is streamlined to focusing on the total enclosed charge by those cylinders within
See lessThree charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 3m
To find the electric flux density (D) at (R = 3m) in the arrangement of charged cylindrical sheets described, we use Gauss's law in its integral form which relates the electric flux through a closed surface to the charge enclosed by that surface. For a cylindrical symmetry, the electric field (E) (aRead more
To find the electric flux density (D) at (R = 3m) in the arrangement of charged cylindrical sheets described, we use Gauss’s law in its integral form which relates the electric flux through a closed surface to the charge enclosed by that surface. For a cylindrical symmetry, the electric field (E) (and thus the electric flux density (D), since (D = epsilon_0 E) in vacuum or air where (epsilon_0) is the permittivity of free space) at a distance (R) from the axis is due to the enclosed charge per unit length ((lambda)) divided by (2piepsilon_0 R).
Given, we have three cylindrical sheets:
– The first with a surface charge density (sigma = 5 , text{C/m}^2) at (R = 2m),
– The second with (sigma = -2 , text{C/m}^2) at (R = 4m),
– The third with (sigma = -3 , text{C/m}^2) at (R = 5m).
Given charges are surface charge densities on cylindrical sheets; thus, the enclosed charge by a Gaussian cylinder of radius (R = 3m) only includes the charge from the first cylindrical sheet ((R = 2m)).
To calculate the total enclosed charge per unit length (
See lessThree charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 1m.
To find the electric flux density (D) at (R = 1m) for the given system of charged cylindrical sheets, we must understand a fundamental concept in electromagnetism. The electric flux density (D) outside a charged cylindrical sheet is determined by the total charge enclosed per unit length divided byRead more
To find the electric flux density (D) at (R = 1m) for the given system of charged cylindrical sheets, we must understand a fundamental concept in electromagnetism. The electric flux density (D) outside a charged cylindrical sheet is determined by the total charge enclosed per unit length divided by (2piepsilon_0), where (epsilon_0) is the vacuum permittivity. Inside the cylindrical sheet or in regions where there is no charge enclosed, the electric flux density due to those charges is zero.
Given:
1. A cylindrical sheet with surface charge density (sigma = 5 , text{C/m}^2) at (R = 2m).
2. A cylindrical sheet with surface charge density (sigma = -2 , text{C/m}^2) at (R = 4m).
3. A cylindrical sheet with surface charge density (sigma = -3 , text{C/m}^2) at (R = 5m).
To find the electric flux density (D) at (R = 1m), we only need to consider the fields generated by charges on the inside surface of our point of interest, which is (R =1m) in this case. Since this point is inside all given cylindrical sheets, none of the sheets contribute to the electric flux density at (R = 1m).
Therefore, the answer
See lessFind the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring
To find the electric potential (V) at a point due to a charged ring, we can use the formula for the potential due to a ring of charge. Given a ring with linear charge density (lambda) (charge per unit length) and radius (R), and a point along the axis perpendicular to the plane of the ring and passiRead more
To find the electric potential (V) at a point due to a charged ring, we can use the formula for the potential due to a ring of charge. Given a ring with linear charge density (lambda) (charge per unit length) and radius (R), and a point along the axis perpendicular to the plane of the ring and passing through its center at a distance (x) from the center of the ring, the electric potential (V) at that point is given by:
[ V = frac{kQ}{sqrt{R^2 + x^2}} ]
where (k) is Coulomb’s constant ((k approx 8.987 times 10^9 , text{Nm}^2/text{C}^2)), (Q) is the total charge on the ring, (R) is the radius of the ring, and (x) is the distance of the point from the center of the ring along the axis. The total charge (Q) can be found from the charge density (lambda) and the circumference of the ring ((Q = lambda cdot 2pi R)).
Given that the linear charge density (lambda = 2) units and the radius of the ring (R = 2) meters, and the point is at a distance of (x = 1) meter from the center of the ring along its axis, let’s calculate the potential
See lessThe potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109 )
The potential difference (V) between two cylinders in a coaxial cable where the inner cylinder is charged, and the outer cylinder is grounded (or vice versa), can be found using Gauss's Law for electric fields and the formula for the electric potential difference.Given:- Charge density (rho = 1) uniRead more
The potential difference (V) between two cylinders in a coaxial cable where the inner cylinder is charged, and the outer cylinder is grounded (or vice versa), can be found using Gauss’s Law for electric fields and the formula for the electric potential difference.
Given:
– Charge density (rho = 1) unit (assuming this is volumetric charge density (rho) in units of (text{C/m}^3), though typically, for coaxial cables, we use linear charge density (lambda) in units of (text{C/m})).
– Inner radius (a = 1) m
– Outer radius (b = 2) m
– The permittivity of free space (varepsilon_0 = 8.854 times 10^{-12} , text{F/m}) (Farads per meter)
For a coaxial cable, the electric field (E) at any point between the cylinders due to the inner charged cylinder can be found by applying Gauss’s Law:
[ E cdot 2pi r cdot L = frac{rho cdot (pi r^2 – pi a^2) cdot L}{varepsilon_0} ]
For (r > a), solving for (E),
[ E = frac{rho (r^2 – a^2)}{2 varepsilon_0 r} ]
See lessA circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?
The net electric flux, (Phi), crossing a surface that encloses a charge is given by Gauss's law. Gauss's law states that the flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space ((epsilon_0)). Mathematically, Gauss's law is expressed as:[Phi = frac{Read more
The net electric flux, (Phi), crossing a surface that encloses a charge is given by Gauss’s law. Gauss’s law states that the flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space ((epsilon_0)). Mathematically, Gauss’s law is expressed as:
[
Phi = frac{Q_{text{enc}}}{epsilon_0}
]
The charge density is given as a function of (phi), (rho_s = 10sinphi), where (rho_s) represents the surface charge density. The total charge enclosed, (Q_{text{enc}}), can be calculated by integrating the charge density over the entire disc. The surface area element for a disc in polar coordinates is (dA = r,dr,dphi), where (r) is the radius and (phi) is the polar angle. However, since the charge density is given as a function of (phi) without dependence on (r), and considering the circular symmetry and the nature of the sine function, we can analyze this considering the charge distribution’s nature over ([0, 2pi]) range of (phi).
For a disc with a radius of 5m, integrating the charge density over the surface should account for the entire area, but the charge density varies with the angular component
See less