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Poisson 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 Del 2 (V)= -ρ/ε, which is the Poisson equation.
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 Del
2
(V)= -ρ/ε, which is the Poisson
equation.
See lessThe function V = e x sin y + z does not satisfy Laplace equation. State True/False.
b Explanation: Grad (V) = e x sin y i + e x cos y j + k. Div(Grad(V)) = e x sin y – e x sin y + 0= 0.Thus Laplacian equation Div(Grad(V)) = 0 is true.
b
Explanation: Grad (V) = e
x
sin y i + e
x cos y j + k. Div(Grad(V)) = e
x
sin y – e
x
sin y + 0=
0.Thus Laplacian equation Div(Grad(V)) = 0 is true.
See lessCalculate the charge density when a potential function x 2 + y 2 + z 2 is in air(in 10-9 order
a Explanation: The Poisson equation is given by Del 2 (V) = -ρ/ε. To find ρ, put ε = 8.854 x 10 -12 in air and Laplacian of the function is 2 + 2 + 2 = 6. Ρ = 6 x 10 -9 /36π = 1/6π units.
a
Explanation: The Poisson equation is given by Del
2
(V) = -ρ/ε. To find ρ, put ε = 8.854 x
10
-12 in air and Laplacian of the function is 2 + 2 + 2 = 6. Ρ = 6 x 10
-9
/36π = 1/6π units.
See lessSuppose 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
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.
See lessIf 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
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.
See lessIn free space, the Poisson equation becomes
c Explanation: The Poisson equation is given by Del 2 (V) = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del 2 (V) = 0, which is the Laplace equation
c
Explanation: The Poisson equation is given by Del
2
(V) = -ρ/ε. In free space, the charges
will be zero. Thus the equation becomes, Del
2
(V) = 0, which is the Laplace equation
See lessThe 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)) = Del 2 (V) = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true
a
Explanation: Grad (V) = 2xi + 2yj – 4zk. Div (Grad (V)) = Del
2
(V) = 2+2-4 = 0. It satisfies
the Laplacian equation. Thus the statement is true
See lessFind 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
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
See lessA 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
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.
See lessThe 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
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.
See less