Electrostatic Boundary value problems Sandra Cruz-Pol, Ph. D. INEL 4151 ch6 Electromagnetics I ECE UPRM Mayagüez, PR

Last Chapters: we knew either V or charge distribution, to find E,D. NOW: Only know values of V or Q at some places (boundaries).

Some applications Microstrip lines capacitance Microstrip lines capacitance Microstrip disk for microwave equipment Microstrip disk for microwave equipment

To find E, we will use: Poisson’s equation: Poisson’s equation: Laplace’s equation: Laplace’s equation: (if charge-free) (if charge-free) They can be derived from Gauss’s Law

Depending on the geometry: We use appropriate coordinates: cartesian:cylindrical:spherical:

Procedure for solving eqs. 1. Choose Laplace (if no charge) or Poisson 2. Solve by Integration if one variable or by 3. Separation of variables if many variables 4. Apply B.C. 5. Find V, then E=-  V, D=  E, J=  E 6. Also, if necessary:

P.E. 6.1 In a 1-dimensional device, the charge density is given by If E=0 at x=0 and V=0 at x=a, find V and E. Evaluating B.C.

45 o P.E. 6.3 two conducting plates of size 1x5m are inclined at 45 o to each other with a gap of width 4mm separating them as shown below. Find approximate charge per plate if plates are kept at 50V potential difference and medium between them has permittivity of 1.5 Applying B.C. V(0)=0, V(  o =45)=V o =50 1m

P.E. 6.3 two conducting plates of size 1x5m are inclined at 45 o to each other with a gap of width 4mm separating them as shown below. permittivity of 1.5 Applying B.C. V(0)=0, V(  o =45)=V o =50 45 o a b

detail a b

P.E. 6.5 Determine the potential function for the region inside the rectangular trough of infinite length whose cross section is shown. The potential V depends on x and y. The potential V depends on x and y. V o =100V, b=2a=2m, find V and E at: V o =100V, b=2a=2m, find V and E at: a) (x, y)=(a, a/2) b) (x, y)=(3a/2, a/4) y x b a V=V o V=0

P.E. 6.5 (cont.) Since it’s 2 variables, use Separation of Variables y x b a V=V o V=0

Let’s examine 3 Possible Cases  =0  <0 C. >0

Case A: If =o y x b a V=V o V=0 This is a trivial solution, therefore cannot be equal to zero.

Case B: <o This is another trivial solution, therefore cannot be equal to zero.

Case C: >o

By superposition, the combination is also a solution:

Cont. B.C. at y=a If we multiply by sin factor and integrate on x: Orthogonality property of sine and cosine:

Flux lines Equipotential lines V=V o V=0

Find V(a,a/2) where V o =100V, b=2a=2m Flux lines Equipotential lines V=V o V=0

Find E at (a,a/2)

Resistance and Capacitance

Resistance If the cross section of a conductor is not uniform we need to integrate: If the cross section of a conductor is not uniform we need to integrate: Solve Laplace eq. to find V Solve Laplace eq. to find V Then find E from its differential Then find E from its differential And substitute in the above equation And substitute in the above equation

P.E. 6.8 find Resistance of disk of radius b and central hole of radius a. a t b

Capacitance Is defined as the ratio of the charge on one of the plates to the potential difference between the plates: Is defined as the ratio of the charge on one of the plates to the potential difference between the plates: Assume Q and find V (Gauss or Coulomb) Assume Q and find V (Gauss or Coulomb) Assume V and find Q (Laplace) Assume V and find Q (Laplace) And substitute E in the equation. And substitute E in the equation.

Capacitance 1. Parallel plate 2. Coaxial 3. Spherical

Parallel plate Capacitor Charge Q and –Q Charge Q and –Q or or Dielectric,  Plate area, S

Coaxial Capacitor Charge +Q & -Q Charge +Q & -Q Dielectric,  Plate area, S c

Spherical Capacitor Charge +Q & -Q Charge +Q & -Q

What is the Earth's charge? The electrical resistivity of the atmosphere decreases with height to an altitude of about 48 kilometres (km), where the resistivity becomes more-or-less constant. This region is known as the electrosphere. There is about a volt (V) potential difference between the Earth's surface and the electrosphere, which gives an average electric field strength of about 6 V/metre (m) throughout the atmosphere. Near the surface, the fine-weather electric field strength is about 100 V/m. The electrical resistivity of the atmosphere decreases with height to an altitude of about 48 kilometres (km), where the resistivity becomes more-or-less constant. This region is known as the electrosphere. There is about a volt (V) potential difference between the Earth's surface and the electrosphere, which gives an average electric field strength of about 6 V/metre (m) throughout the atmosphere. Near the surface, the fine-weather electric field strength is about 100 V/m. The Earth is electrically charged and acts as a spherical capacitor. The Earth has a net negative charge of about a million coulombs, while an equal and positive charge resides in the atmosphere.

Capacitors connection Series Series Parallel Parallel

Resistance Recall that: Recall that: Multiplying, we obtain the Relaxation Time: Multiplying, we obtain the Relaxation Time: Solving for R, we obtain it in terms of C: Solving for R, we obtain it in terms of C:

So In summary we obtained : Capacitor C R=  C Parallel Plate Coaxial Spherical

P.E. 6.9 A coaxial cable contains an insulating material of  1 in its upper half and another material with  2 in its lower half. Radius of central wire is a and of the sheath is b. Find the leakage resistance of length L. A coaxial cable contains an insulating material of  1 in its upper half and another material with  2 in its lower half. Radius of central wire is a and of the sheath is b. Find the leakage resistance of length L. 11 22 They are connected in parallel

P.E. 6.10a Two concentric spherical capacitors with  1r =2.5 in its outer half and another material with  2r =3.5 in its inner half. The inner radius is a=1mm, b=3mm and c=2mm. Find their C. Two concentric spherical capacitors with  1r =2.5 in its outer half and another material with  2r =3.5 in its inner half. The inner radius is a=1mm, b=3mm and c=2mm. Find their C. 11 22 c We have two capacitors in series:

P.E. 6.10b Two spherical capacitors with  1r =2.5 in its upper half and another material with  2r =3.5 in its lower half. Inner radius is a=1mm and b=3mm. Find their C. Two spherical capacitors with  1r =2.5 in its upper half and another material with  2r =3.5 in its lower half. Inner radius is a=1mm and b=3mm. Find their C. 11 22 We have two capacitors in parallel:

Method of Images Whenever the is a charge in the presence of a conductor. The conductor serves as a mirror. Whenever the is a charge in the presence of a conductor. The conductor serves as a mirror. Substitute the conductor for a plane at V=0 and the image. Substitute the conductor for a plane at V=0 and the image. The solution will be valid only for the region above the conductor. The solution will be valid only for the region above the conductor.

Line charge above ground plane