Correct Answer - Option 3 : 0.5
Concept:
Charge density
Pv = ∇ ⋅ D (Gauss-Divergence theorem)
In integral form
ϕ D⋅ds = ∫ρvdv
In cylindrical co-i
\(\nabla \cdot \vec D = \;\frac{1}{\rho}\frac{{\partial \left( {\rho{D_r}} \right)}}{{\partial \rho}} + \frac{1}{\rho}\frac{{\partial {D_\phi }}}{{\partial \phi }} + \frac{{\partial {D_z}}}{{\partial z}}\)
Calculation:
Given D = z ρ (cos2 ϕ) âz
So \(\nabla \cdot D = \frac{\partial }{{\partial z}}\left[ {z\rho {{\cos }^2}\phi } \right]\) (Z-component)
\(\nabla \cdot D = \rho {\cos ^2}\phi = {\rho _v}\)
At \(P = 1,\;\phi = \frac{\pi }{4}\)
\({\rho _v} = 1 \cdot {\cos ^2}\left( {\frac{\pi }{4}} \right) = \frac{1}{2}\)
So ρv = 0.5 c/m3
Option (c) correct choice.
More Information:
Maxwell equation for static and magnetic field
Differential (or point form)
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Integral form
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Remark
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\(\vec \nabla \cdot \vec D = {\rho _v}\)
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\(\mathop \oint \nolimits_s D \cdot ds = \mathop \smallint \nolimits_v {\rho _v}dv\)
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Gauss law
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\(\vec \nabla \cdot \vec B = 0\)
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\(\mathop \oint \nolimits_s \vec B \cdot \vec ds = 0\)
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Non existence of magnetic monopole
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\(\nabla \times \vec E = 0\)
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\(\mathop \oint \nolimits_L E \cdot dl = 0\)
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Conservative nature of the electrostatic field.
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\(\vec \nabla \times \vec H = \vec J\)
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\(\mathop \oint \nolimits_L H \cdot dl = \mathop \smallint \nolimits_s \vec J \cdot d\vec s\)
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Ampers law.
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Divergence in:
I) Cartesian form: If \(\vec A = {A_x}{q_x} + {A_y}{a_y} + {a_z}{a_z}\)
\(\nabla \cdot \vec A = \frac{2}{{2x}}{A_x} + \frac{2}{{2y}}{A_y} + \frac{\partial }{{\partial z}}{A_z}\)
II) In polar form:
\(\nabla \cdot \vec A = \frac{1}{{{r^2}}}\frac{\partial }{{\partial r}}\left( {{r^2}Ar} \right) + \frac{1}{{r\sin \theta }}\frac{\partial }{{\partial \theta }}(\sin \theta \;A\theta ) + \frac{1}{{r\sin \theta }}\frac{{\partial A}}{{\partial \phi }}\)