There is a polarization vector P in the object whose boundaries are given below. Accordingly, find the equivalent charge density at point K. There is a polarization vector P in the object whose boundaries are given below., find the equivalent charge density at point K object boundaries/object boundary: 0

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There is a polarization vector P in the object whose boundaries are given below. Accordingly, find the equivalent charge density at point K.
There is a polarization vector P in the object whose boundaries are given below., find the equivalent charge density at point K
nC
nC
object boundaries /object boundary: 0 <R<5 m, 0 <0 < 60°,0<Ø< 2π
m³.
P=75ax+ 5az
K(2.5; 30°; 180°)
x=rcos (0) = Rsin(8)cos(0)
y=rsin(0) = Rsin(8)sin(0)
z = Rcos(8)
â, = cos (0)â+ sin (0)â,
â = -sin (0)â, + cos (0)â,
a
a
a
V=—â₂ + â, + -â
ax dy Əz
R=√x² + y² +2²
8 = cos-1
Z
√x² + y² + z²
â = sin (8)cos (Ø)â, +sin (8)sin (0)â, + cos (0)â,
âg=cos (8)cos (0)â, + cos (8)sin (0)â, -sin (8)â,
O a. 0
O b. 6.4952
O c. 3,9952
O D. none of these
O to. -0.5801
O f. 4.3301
Transcribed Image Text:There is a polarization vector P in the object whose boundaries are given below. Accordingly, find the equivalent charge density at point K. There is a polarization vector P in the object whose boundaries are given below., find the equivalent charge density at point K nC nC object boundaries /object boundary: 0 <R<5 m, 0 <0 < 60°,0<Ø< 2π m³. P=75ax+ 5az K(2.5; 30°; 180°) x=rcos (0) = Rsin(8)cos(0) y=rsin(0) = Rsin(8)sin(0) z = Rcos(8) â, = cos (0)â+ sin (0)â, â = -sin (0)â, + cos (0)â, a a a V=—â₂ + â, + -â ax dy Əz R=√x² + y² +2² 8 = cos-1 Z √x² + y² + z² â = sin (8)cos (Ø)â, +sin (8)sin (0)â, + cos (0)â, âg=cos (8)cos (0)â, + cos (8)sin (0)â, -sin (8)â, O a. 0 O b. 6.4952 O c. 3,9952 O D. none of these O to. -0.5801 O f. 4.3301
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