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When a material has zero permittivity, the maximum potential that it can possess is
d Explanation: Permittivity is zero, implies that the ability of the material to store electric charges is zero. Thus the electric field and potential of the material is also zero.
d
See lessExplanation: Permittivity is zero, implies that the ability of the material to store electric charges is zero. Thus the electric field and potential of the material is also zero.
Find the charge density from the function of flux density given by 12x – 7z.
c Explanation: From point form of Gauss law, we get Div (D) = ρv Div (D) = Div(12x – 7z) = 12-7 = 5, which the charge density ρv. Thus ρv = 5 units.
c
Explanation: From point form of Gauss law, we get Div (D) = ρv
Div (D) = Div(12x – 7z) = 12-7 = 5, which the charge density ρv. Thus ρv = 5 units.
See lessPoisson equation can be derived from which of the following equations?
a Explanation: The point of Gauss law is given by, Div (D)= ρv. On putting D= ε E and E=- Grad (V) in Gauss law, we get Del2 (V)= -ρ/ε, which is the Poisson equation.
a
See lessExplanation: The point of Gauss law is given by, Div (D)= ρv. On putting D= ε E and E=- Grad (V) in Gauss law, we get Del2 (V)= -ρ/ε, which is the Poisson equation.
Suppose the potential function is a step function. The equation that gets satisfied is
a Explanation: Step is a constant function. The Laplace equation Div(Grad(step)) will become zero. This is because gradient of a constant is zero and divergence of zero vector will be zero.
a
See lessExplanation: Step is a constant function. The Laplace equation Div(Grad(step)) will
become zero. This is because gradient of a constant is zero and divergence of zero
vector will be zero.
If Laplace equation satisfies, then which of the following statements will be true?
b Explanation: Laplace equation satisfying implies the potential is not necessarily zero due to subsequent gradient and divergence operations following. Finally, if potential is assumed to be zero, then resistance is zero and current will be infinite.
b
See lessExplanation: Laplace equation satisfying implies the potential is not necessarily zero due to subsequent gradient and divergence operations following. Finally, if potential is assumed to be zero, then resistance is zero and current will be infinite.
In free space, the Poisson equation becomes
c Explanation: The Poisson equation is given by Del2 (V) = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del2(V) = 0, which is the Laplace equation.
c
See lessExplanation: The Poisson equation is given by Del2 (V) = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del2(V) = 0, which is the Laplace equation.
The given equation satisfies the Laplace equation. V = x 2 + y 2 – z 2 . State True/False.
a Explanation: Grad (V) = 2xi + 2yj – 4zk. Div (Grad (V)) = Del2 (V) = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true.
a
See lessExplanation: Grad (V) = 2xi + 2yj – 4zk. Div (Grad (V)) = Del2
(V) = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true.
Find the permittivity of the surface when a wave incident at an angle 60 is reflected by the surface at 45 in air.
d Explanation: From the relations of the boundary conditions of a dielectric-dielectric interface, we get tan θ1/tan θ2 = ε1/ε2. Thus tan 60/tan 45 = ε1/1. We get ε1 = tan 60 =1.73.
d
See lessExplanation: From the relations of the boundary conditions of a dielectric-dielectric
interface, we get tan θ1/tan θ2 = ε1/ε2. Thus tan 60/tan 45 = ε1/1. We get ε1 = tan 60 =1.73.
A wave incident on a surface at an angle 60 degree is having field intensity of 6 units. The reflected wave is at an angle of 30 degree. Find the field intensity after reflection.
c Explanation: By Snell’s law, the relation between incident and reflected waves is given by, E1 sin θ1 = E2 sin θ2. Thus 6 sin 60 = E2 sin 30. We get E2 = 6 x 1.732 = 10.4 units.
c
See lessExplanation: By Snell’s law, the relation between incident and reflected waves is given
by, E1 sin θ1 = E2 sin θ2. Thus 6 sin 60 = E2 sin 30. We get E2 = 6 x 1.732 = 10.4 units.
The electric field intensity of a surface with permittivity 3.5 is given by 18 units. What the field intensity of the surface in air?
c Explanation: The relation between flux density and permittivity is given by En1/En2 = ε2/ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.
c
See lessExplanation: The relation between flux density and permittivity is given by En1/En2 = ε2/ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.