EEE340Lecture 141 Example 3-14 A lucite sheet Determine Solution. Boundary condition Eq. (3-123) Figure 3-24 E0E0 EiEi E0E0

EEE340Lecture 142 Example 3-15 Determine the magnitude and direction of the electric field at point P 2, giving E 1 at P 1 with Solution Figure 3-25 Question: which is larger, E 1n E 1t E 2n E 2t

EEE340Lecture 143 Example 3-16: Cylindrical Cable. Given r i =0.4 cm,  rr =3.2,  rp =2.6, r p =0.616 cm, r p =0.832 cm, Find the E-field profile. Solution: Apply the Gaussian law. i.e. In rubber In polystyrene riri rprp roro

EEE340Lecture 144 In rubber: In polystyrene:

EEE340Lecture 145 or riri rprp roro

EEE340Lecture : Capacitance and Capacitors The capacitance of a capacitor is a physical property of the two-conductor system: For a parallel plate capacitor Note that in (3.135) Q is from one conductor (with positive charge). Hence (3.135) (3.122) (3.136)

EEE340Lecture 147 Example 3-18: Cylindrical capacitor with length L, radii a and b, and permittivity . Solution: Using Gauss’s law Hence (3.138) (3.139)

EEE340Lecture 148 Example3-19: Spherical capacitor with r i and r o Solution From Gauss’s law Hence The earth,

EEE340Lecture 149 Example: Capacitors in series A coaxial cable is partially filled with a dielectric  r 1 =10 and air  r 2 =1. The dimensions are  1 =1cm,  2 =2cm,  3 =3cm Find the capacitance per unit length Solution Apply Gauss’s law Therefore Where Q is the charge per unit length.

EEE340Lecture 1410 Per unit length capacitance