Explanation Given: The range of conical section of a material is 2.0 cm ≤ r ≤ 9.0 cm and 0 ≤ θ ≤ 30 ∘ . The relative permittivity ε r is 9.0 and the conductivity σ is 0.020 S / m . Concept used: The Laplace equation for spherical coordinates is shown below. ∇ 2 V spherical = 1 r 2 ∂ ∂ r ( r 2 ∂ V ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ V ∂ θ ) + 1 r 2 sin 2 θ d 2 V d ϕ 2 Calculation: Consider the inner radius be a and outer radius be b for the conical section of a material. The potential is a function of r so, Laplace equation can be written as, ∇ 2 V = 0 1 r 2 ∂ ∂ r ( r 2 ∂ V ∂ r ) = 0 ∂ ∂ r ( r 2 ∂ V ∂ r ) = 0 Integrate the above equation with respect to r . r 2 ∂ V ∂ r = A ∂ V ∂ r = A r 2 Integrate the above equation with respect to r . V ( r ) = ∫ A r 2 d r = A ( − 1 r ) + B = − A r + B Here, A and B are constants of integration and can be obtained from boundary conditions. The first boundary condition is V ( a ) = 0 , since the potential at outer surface is zero. So, the constant can be obtained as, 0 = − A a + B B = A a Substitute A a for B in the expression for V ( r ) as, V ( r ) = − A r + A a = − A ( 1 r − 1 a ) The second boundary condition is V ( b ) = V r . So, substitute V r for V ( r ) in above equation. V r = − A ( 1 b − 1 a ) A = V r ( 1 a − 1 b ) Now, the expression for V ( r ) can be written with the values of constants known as, V ( r ) = − V r ( 1 a − 1 b ) ( 1 r − 1 a ) = V r ( 1 a − 1 r ) ( 1 a − 1 b ) The electric field intensity E is calculated as, E = − ∇ V ( r ) = − ∂ V ∂ r a r = − ∂ ∂ ρ ( V r ( 1 a − 1 r ) ( 1 a − 1 b ) ) a r = − V r ( 1 a − 1 b ) ∂ ∂ ρ [ ( 1 a − 1 r ) ] a r Simplify the above equation. E = − V r ( 1 a − 1 b ) ( 1 r 2 ) a r = − V r r 2 ( 1 a − 1 b ) a r The electric flux density can be written as, D = ε r ε 0 E = ε r ε 0 ( − V r r 2 ( 1 a − 1 b ) ) a r = − ε r ε 0 V r r 2 ( 1 a − 1 b ) a r The surface charge density ρ s at the conductive plates is, ρ s = D | r = b = − ε r ε 0 V r b 2 ( 1 a − 1 b ) The expression for charge Q on inner surface is shown below

Fundamentals of Electromagnetics with Engineering Applications

1st Edition

ISBN: 9780470105757

Author: Stuart M. Wentworth

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2, Problem 2.72P

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The resistance and capacitance of conical section of material.

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Determine the capacitance of the given spherical cross section as shown in figure belowwith the following specifications: a =1 mm, b = 3 mm, c= 2 m, εr 1=2.5∧εr2=3.5.

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The resistance and capacitance of conical section of material.

Solution Summary: The author calculates the resistance and capacitance of the conical section of a material. The Laplace equation for spherical coordinates is shown.

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