Question

Given that the electric flux density 

D = zρ (cos² Φ) âz, C/m² The charge density at point (1,π/4, 3) is

1. 3
2. 1
3. 0.5
4. 0.5 a

Answer 1

Correct Answer - Option 3 : 0.5

Concept:

Charge density

P_v = ∇ ⋅ D (Gauss-Divergence theorem)

In integral form

ϕ D⋅ds = ∫ρ_vdv

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) â_z

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 ρ_v = 0.5 c/m³

Option (c) correct choice.

More Information:

Maxwell equation for static and magnetic field

Differential (or point form)	Integral form	Remark
\(\vec \nabla \cdot \vec D = {\rho _v}\)	\(\mathop \oint \nolimits_s D \cdot ds = \mathop \smallint \nolimits_v {\rho _v}dv\)	Gauss law
\(\vec \nabla \cdot \vec B = 0\)	\(\mathop \oint \nolimits_s \vec B \cdot \vec ds = 0\)	Non existence of magnetic monopole
\(\nabla \times \vec E = 0\)	\(\mathop \oint \nolimits_L E \cdot dl = 0\)	Conservative nature of the electrostatic field.
\(\vec \nabla \times \vec H = \vec J\)	\(\mathop \oint \nolimits_L H \cdot dl = \mathop \smallint \nolimits_s \vec J \cdot d\vec s\)	Ampers law.

 

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:

\(\nabla \cdot \vec A = \frac{1}{{{r^2}}}\frac{\partial }{{\partial r}}\left( {{r^2}Ar} \right) + \frac{1}{{r\sin \theta }}\frac{\partial }{{\partial \theta }}(\sin \theta \;A\theta ) + \frac{1}{{r\sin \theta }}\frac{{\partial A}}{{\partial \phi }}\)