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If a potential V is 2V at x = 1mm and is zero at x=0 and volume charge density is -106εo, constant throughout the free space region between x = 0 and x = 1mm. Calculate V at x = 0.5mm.
Answer: d Explanation: Del2(V) = -ρv/εo= +106 On integrating twice with respect to x, V = 106. (x2/2) + C1x + C2. Substitute the boundary conditions, x = 0, V = 0 and x = 1mm, V = 2V in V, C1 = 1500 and C2 = 0. At x = 0.5mm, we get, V = 0.875V.
Answer: d
Explanation: Del2(V) = -ρv/εo= +106
On integrating twice with respect to x, V = 106. (x2/2) + C1x + C2.
Substitute the boundary conditions, x = 0, V = 0 and x = 1mm, V = 2V in V,
See lessC1 = 1500 and C2 = 0. At x = 0.5mm, we get, V = 0.875V.
Given the potential V = 25 sin θ, in free space, determine whether V satisfies Laplace’s equation
Answer: a Explanation: (Del)2V = 0 (Del)2V = (Del)2(25 sin θ), which is not equal to zero. Thus the field does not satisfy Laplace equation.
Answer: a
See lessExplanation: (Del)2V = 0
(Del)2V = (Del)2(25 sin θ), which is not equal to zero. Thus the field does not satisfy
Laplace equation.
The Poisson equation cannot be determined from Laplace equation. State True/False.
Answer: b Explanation: The Poisson equation is a general case for Laplace equation. If volume charge density exists for a field, then (Del)2V= -ρv/ε, which is called Poisson equation.
Answer: b
See lessExplanation: The Poisson equation is a general case for Laplace equation. If volume
charge density exists for a field, then (Del)2V= -ρv/ε, which is called Poisson equation.
If a function is said to be harmonic, then
Answer: c Explanation: Though option Curl(Grad V) = 0 & Div(Curl V) = 0 are also correct, for harmonic fields, the Laplacian of electric potential is zero. Now, Laplacian refers to Div(Grad V), which is zero for harmonic fields.
Answer: c
See lessExplanation: Though option Curl(Grad V) = 0 & Div(Curl V) = 0 are also correct, for
harmonic fields, the Laplacian of electric potential is zero. Now, Laplacian refers to
Div(Grad V), which is zero for harmonic fields.
The point form of Gauss law is given by, Div(V) = ρv State True/False.
Answer: a Explanation: The integral form of Gauss law is ∫∫∫ ρv dv = V. Thus differential or point form will be Div(V) = ρv.
Answer: a
See lessExplanation: The integral form of Gauss law is ∫∫∫ ρv dv = V. Thus differential or point form will be Div(V) = ρv.
Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.
Answer: b Explanation: Div (D) = 2y ∫∫∫Div (D) dv = ∫∫∫ 2y dx dy dz. On integrating, x = 0->1, y = 0->2 and z = 0->3, we get Q =12.
Answer: b
See lessExplanation: Div (D) = 2y
∫∫∫Div (D) dv = ∫∫∫ 2y dx dy dz. On integrating, x = 0->1, y = 0->2 and z = 0->3, we get Q =12.
Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y3 /3 j.
Answer: c Explanation: Div(D) = 10y2 ∫∫∫Div (D) dv = ∫∫∫ 10y2 dx dy dz. On integrating, x = -1->1, y = -1->1 and z = -1->1, we get Q = 80/3.
Answer: c
See lessExplanation: Div(D) = 10y2
∫∫∫Div (D) dv = ∫∫∫ 10y2 dx dy dz. On integrating, x = -1->1, y = -1->1 and z = -1->1, we get
Q = 80/3.
Using volume integral, which quantity can be calculated?
Answer: c Explanation: The volume integral gives the volume of a vector in a region. Thus volume of a cube can be computed.
Answer: c
See lessExplanation: The volume integral gives the volume of a vector in a region. Thus volume of a cube can be computed.
Compute the Gauss law for D = 10ρ3 /4 i, in cylindrical coordinates with ρ = 4m, z = 0 and z = 5, hence find charge using volume integral.
Answer: d Explanation: Q = D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed. The divergence of D given is, Div(D) = 10 ρ2 and dv = ρ dρ dφ dz. On integrating, ρ = 0->4, φ = 0->2π and z = 0->5, we get Q = 6400 π.
Answer: d
See lessExplanation: Q = D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 10 ρ2 and dv = ρ dρ dφ dz. On integrating, ρ = 0->4, φ = 0->2π and z = 0->5, we get Q = 6400 π.
Compute divergence theorem for D = 5r2 /4 i in spherical coordinates between r = 1 and r = 2 in volume integral.
Answer: c Explanation: D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed. The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r =1->2, φ = 0->2π and θ = 0->π, we get Q = 75 π.
Answer: c
See lessExplanation: D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r =1->2, φ = 0->2π and θ = 0->π, we get Q = 75 π.