1. ENE/EIE 325 Electromagnetic Fields and Waves
Lecture 7
Boundary Conditions and Capacitance

2. Ex1 The isotropic dielectric medium with r1 = 3 and r2 = 2 is connected as shown. Given V/m, determine and its magnitude, and its magnitude, 1, and 2.

3. Solution: Use Boundary Conditions. Since boundary is X-Y plane
Therefore, any vector that lies parallel to this plane is considered
“tangential”.
Since:
Apply BC’s:
Hence:
Apply BC’s: or
and so

4. Therefore
Next
C/m2
where o = x F/m

5. Ex2 Between a dielectric-conductor interface has a surface charge density of s = 2x109 C/m2. Given V/m, determine

6. Capacitance
Capacitance depends on the shape of conductor and the permittivity of the medium. Capacitance has a unit of Farad or F.
From
then

7. Capacitance for parallel plate configuration
At lower plate,
then
The potential difference
Let A = the plate area then Q = sA
then

8. Total energy stored in the capacitance

9. Ex3 Determine the relative permittivity of the dielectric material inserted between a parallel plate capacitor if
C = 40 nF, d = 0.1 mm, and A = 0.15 m2
d = 0.2 mm, E = 500 kV/m, and s = 10 C/m2

10. Capacitance in various charge distribution configurations (1)
Coaxial cable
Use Gauss’s law,
…and the right hand side becomes:

11. Farads

12. Capacitance in various charge distribution configurations (2)
Sphere
Use Gauss’s law,

13. Farads

14. Capacitance in various charge distribution configurations (3)
A parallel plate capacitor with horizontal dielectric layers
Solution: suppose we assume a potential difference Vo between the plates. The electric field intensities between the two regions: E2 and E1 are both uniform and
At the dielectric interface:
Eliminate E2 in our Vo relation:

15. And the surface charge density on the lower plate therefore has the magnitude:
Because D1 = D2, the magnitude of the surface charge is the same on each plate. The capacitance is then
The two capacitors are connected in series!

16. Capacitance in various charge distribution configurations (4)
A parallel plate capacitor with vertical dielectric layers
Assume a potential difference Vo would
produce field strength E1 = E2 = Vo/d.
These are tangential fields at the interface,
and they must be equal.

17. The two capacitors are in parallel.

18. Ex4 From the parallel capacitor shown, find the total capacitance.