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

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.

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

Therefore Next C/m2 where o = 8.8254 x 10-12 F/m

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

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

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

Total energy stored in the capacitance

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

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

Farads

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

Farads

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:

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!

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.

The two capacitors are in parallel.

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