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f 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
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
d
See lessExplanation: 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
Given the potential V = 25 sin θ, in free space, determine whether V satisfies Laplace’s equation
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.
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.
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
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.
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.
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
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.
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 integra
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
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
If D = 2xy i + 3yz j + 4xz k, how much flux passes through x = 3 plane for which -1<y<2 and 0<z<4?
c Explanation: By Gauss law, ψ = ∫∫ D.ds, where ds = dydz i at the x-plane. Put x = 3 and integrate at -1<y<2 and 0<z<4, we get 12 X 3 = 36.
c
See lessExplanation: By Gauss law, ψ = ∫∫ D.ds, where ds = dydz i at the x-plane. Put x = 3 and
integrate at -1<y<2 and 0<z<4, we get 12 X 3 = 36.
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.
b Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
b
See lessExplanation: While evaluating surface integral, there has to be two variables and one
constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx
dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
Find the value of divergence theorem for A = xy2 i + y3 j + y2z k for a cuboid given by 0<x<1, 0<y<1 and 0<z<1
c Explanation: A cuboid has six faces. ∫∫A.ds = ∫∫Ax=0 dy dz + ∫∫Ax=1 dy dz + ∫∫Ay=0 dx dz + ∫∫Ay=1 dx dz + ∫∫Az=0 dy dx + ∫∫Az=1 dy dx. Substituting A and integrating we get (1/3) + 1 + (1/3) = 5/3
c
See lessExplanation: A cuboid has six faces. ∫∫A.ds = ∫∫Ax=0 dy dz + ∫∫Ax=1 dy dz + ∫∫Ay=0 dx dz
+ ∫∫Ay=1 dx dz + ∫∫Az=0 dy dx + ∫∫Az=1 dy dx. Substituting A and integrating we get (1/3)
+ 1 + (1/3) = 5/3
Compute divergence theorem for D= 5r2 /4 i in spherical coordinates between r=1 and r=2
c Explanation: ∫∫ ( 5r2 /4) . (r2 sin θ dθ dφ), which is the integral to be evaluated. Since it is double integral, we need to keep only two variables and one constant compulsorily. Evaluate it as two integrals keeping r = 1 for the first integral and r = 2 for the second integral, with φ = 0→2π andRead more
c
See lessExplanation: ∫∫ ( 5r2
/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated. Since it is
double integral, we need to keep only two variables and one constant compulsorily.
Evaluate it as two integrals keeping r = 1 for the first integral and r = 2 for the second
integral, with φ = 0→2π and θ = 0→ π. The first integral value is 80π, whereas second
integral gives -5π. On summing both integrals, we get 75π
Compute the Gauss law for D= 10ρ3 /4 i, in cylindrical coordinates with ρ= 4m, z=0 and z=5
d Explanation: ∫∫ D.ds = ∫∫ (10ρ3 /4).(ρ dφ dz), which is the integral to be evaluated. Put ρ = 4m, z = 0→5 and φ = 0→2π, the integral evaluates to 6400π
d
See lessExplanation: ∫∫ D.ds = ∫∫ (10ρ3
/4).(ρ dφ dz), which is the integral to be evaluated. Put ρ =
4m, z = 0→5 and φ = 0→2π, the integral evaluates to 6400π