Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
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 = 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 lessWhat is the potential difference between 10sinθcosφ/r 2 at A(1,30,20) and B(4,90,60)?
To solve this question and find the potential difference between the two points, we'll interpret the potential function as (V(x,y,z) = frac{10sinthetacosphi}{r^2}), where (r) is the distance from the point of reference, usually the origin, (x) is represented as (rsinthetacosphi), (y) as (rsinthetasiRead more
To solve this question and find the potential difference between the two points, we’ll interpret the potential function as (V(x,y,z) = frac{10sinthetacosphi}{r^2}), where (r) is the distance from the point of reference, usually the origin, (x) is represented as (rsinthetacosphi), (y) as (rsinthetasinphi), and (z) as (rcostheta). These are the spherical coordinates transformation equations. Given points A and B are provided in spherical coordinates as (A(r,theta,phi) = A(1, 30°, 20°)) and (B(r,theta,phi) = B(4, 90°, 60°)).
First, let’s evaluate the potential at point A ((V_A)):
– (r_A = 1)
– (theta_A = 30°)
– (phi_A = 20°)
[V_A = frac{10sin(30°)cos(20°)}{1^2}]
To convert the angles into radians which is often the required format for mathematical functions in calculators and programming languages:
– (30° = frac{pi}{6}) radians and (20° = frac{pi}{9}) radians approximately.
[
See lessGiven E = 40xyi + 20×2 j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3)
To find the potential difference between two points given an electric field vector ( mathbf{E} = 40xymathbf{i} + 20x^2 mathbf{j} + 2mathbf{k} ), we recognize this requires integrating the electric field along a path from point A ((1, -1, 0)) to point B ((2, 1, 3)). The potential difference ((V)) betRead more
To find the potential difference between two points given an electric field vector ( mathbf{E} = 40xymathbf{i} + 20x^2 mathbf{j} + 2mathbf{k} ), we recognize this requires integrating the electric field along a path from point A ((1, -1, 0)) to point B ((2, 1, 3)). The potential difference ((V)) between two points in an electric field is found using the negative integral of the electric field along the path from point A to point B. Mathematically, this is given as:
[ V = – int_{A}^{B} mathbf{E} cdot dmathbf{r} ]
Given ( mathbf{E} = 40xymathbf{i} + 20x^2mathbf{j} + 2mathbf{k} ), let’s decompose this problem into a manageable form.
The differential length vector ( dmathbf{r} ) in Cartesian coordinates is given by
[ dmathbf{r} = dxmathbf{i} + dymathbf{j} + dzmathbf{k} ]
So, the dot product ( mathbf{E} cdot dmathbf{r} ) becomes
[ mathbf{E} cdot dmath
See lessA point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge
The electric potential (V) due to a point charge (q) at a distance (r) in a vacuum is given by the formula:[ V = frac{1}{4piepsilon_0} cdot frac{q}{r} ]where (epsilon_0) is the vacuum permittivity constant, approximately equal to (8.85 times 10^{-12} , text{C}^2/text{N}cdottext{m}^2).The potential dRead more
The electric potential (V) due to a point charge (q) at a distance (r) in a vacuum is given by the formula:
[ V = frac{1}{4piepsilon_0} cdot frac{q}{r} ]
where (epsilon_0) is the vacuum permittivity constant, approximately equal to (8.85 times 10^{-12} , text{C}^2/text{N}cdottext{m}^2).
The potential difference (Delta V) between two points due to a point charge is the difference in the electric potentials at those two points, given by:
[ Delta V = V_2 – V_1 ]
Considering the point charge of (0.4 text{nC}) (or (0.4 times 10^{-9} , text{C})) located at ((2, 3, 3)) meters, we want to find the potential difference between points ((2, 3, 3)) meters and ((-2, 3, 3)) meters.
### Calculation
1. Distance of the first point from the charge:
The first point is the location of the charge itself, so (r_1 = 0).
– For practical purposes, the potential at the location of a point charge is infinite, but since we’re calculating a
See lessSix equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.
To find the electric potential at the origin due to six point charges, (Q = 10 , text{nC} = 10 times 10^{-9} , text{C}), located at distances of 2, 3, 4, 5, 6, and 7m, we can use the formula:[V = frac{1}{4piepsilon_0} sum frac{Q}{r_i}]Where:- (V) is the electric potential,- (Q) is the charge,- (r_i)Read more
To find the electric potential at the origin due to six point charges, (Q = 10 , text{nC} = 10 times 10^{-9} , text{C}), located at distances of 2, 3, 4, 5, 6, and 7m, we can use the formula:
[V = frac{1}{4piepsilon_0} sum frac{Q}{r_i}]
Where:
– (V) is the electric potential,
– (Q) is the charge,
– (r_i) is the distance of each charge from the point where the potential is being calculated (in this case, the origin),
– (epsilon_0) is the permittivity of free space ((8.85 times 10^{-12} , text{F/m})).
Plugging in the values:
[V = frac{1}{4pi(8.85 times 10^{-12})} left( frac{10 times 10^{-9}}{2} + frac{10 times 10^{-9}}{3} + frac{10 times 10^{-9}}{4} + frac{10 times 10^{-9}}{5} + frac{10 times 10^{-9}}{6} + frac{10 times 10
See lessFind the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109 units)
The electric field intensity (E) in a medium, such as transformer oil, with a relative permittivity (εr) is not directly determined by the relative permittivity or the density alone. Instead, the electric field in a region is determined by the presence of electric charges and the distribution of theRead more
The electric field intensity (E) in a medium, such as transformer oil, with a relative permittivity (εr) is not directly determined by the relative permittivity or the density alone. Instead, the electric field in a region is determined by the presence of electric charges and the distribution of these charges in and around that medium.
However, to provide context to your query, we could interpret it as seeking the electric field (E) in a scenario where a certain electric potential or voltage is applied across a medium, like transformer oil, assuming no free charges are present within the oil itself. In a vacuum or free space, the electric field (E) due to a point charge is given by Coulomb’s law, (E = frac{k cdot q}{r^2}), where (k) is Coulomb’s constant ((8.987 times 10^9 N m^2/C^2)), (q) is the charge in Coulombs, and (r) is the distance from the charge in meters.
In a medium like transformer oil, this field is modified by the material’s relative permittivity ((ε_r)), which is a measure of how an electric field within the material is reduced compared to the field in a vacuum. The electric field in a material is thus given more generally by (E = frac{k cdot q}{ε_0 cdot ε_r cdot r^2}), where
See less