Dielectrics

1. Dielectrics • Conductor has free electrons. • Dielectric electrons are strongly bounded to the atom. • In a dielectric, an externally applied electric field, Eext cannot cause mass migration of charges since none are able to move freely. • But, Eext can polarize the atoms or molecules in the material. • The polarization is represented by an electric dipole.

2. Note the field will apply a force on both the positively charged nucleus and the negatively charged electron. However, these forces will move these particles in opposite directions • Note, an electric dipole has been created !

3. D, flux density is proportionally increase as polarization increase through induction of permittivity, ε of the material relating the E and D permittivity, ε = proportional to the permittivity of free space, ε0 ε = εrε0

4. However, the electron may be break free from the atom, creating a positive ion and a free electron. • We call these free charges, and the electric field will cause them to move in opposite directions : • Moving charge is electric current J(r ) .

5. Electric Boundary Conditions • Electric field maybe continuous in each of two dissimilar media • But, the E-field maybe discontinuous at the boundary between them • Boundary conditions specify how the tangential and normal components of the field in one medium are related to the components in other medium across the boundary • Two dissimilar media could be: two different dielectrics, or a conductor and a dielectric, or two conductors

6. Dielectric- dielectric boundary • Interface between two dielectric media

7. Dielectric- dielectric boundary • Based on the figure in previous slide: • First boundary condition related to the tangential components of the electric field E is: • Second boundary condition related to the normal components of the electric field E is: • OR

8. z ε1 E2 Ez2 Exy1 Exy2 xy -plane ε2 Ez1 E1

9. xy • Solution: • Exy1=Exy2 thus,Exy2 = 3ax+4ay • Ez1 = 5az, • but, Ez2 = ?? • 2ε0(5az) = 5ε0(Ez2) • Ez2 = 2az • thus, E2 =3az+4ay+2az ε1=2ε0 ε2=5ε0 E2 Example 1: 1) Find E2 in the dielectric, when E1 = 3ax+4ay+5az, 2) And find Ө1 and Ө2. Ө2 z Ө1 E1

10. z ε1=2ε0 Find E1 if E2 = 2x -3y +3z with s = 3.54 x 10-11(C/m2) And find Ө1 and Ө2 Ө2 E2 xy ε2=8ε0 E1 Ө1

11. Conductor- conductor boundary • Boundary between two conducting media: • Using the 1st and 2nd boundary conditions: and

12. xy J2 Jxy2 Jz1 Jz2 z Jxy1 J1

13. Conductor- conductor boundary • In conducting media, electric fields give rise to current densities. • From , we have: and • The normal component of J has be continuous across the boundary between two different media under electrostatic conditions.

14. Conductor- conductor boundary • Hence, upon setting , we found the boundary condition for conductor- conductor boundary:

15. Dielectric-conductor boundary • Assume medium 1 is a dielectric • Medium 2 is a perfect conductor

16. Perfect conductor • When a conducting slab is placed in an external electric field, • Charges that accumulate on the conductor surfaces induces an internal electric field • Hence, total field inside conductor is zero.

17. Dielectric-conductor boundary • The fields in the dielectric medium, at the boundary with the conductor is . • Since , it follows that . • Using the equation, , • we get: • Hence, boundary condition at conductor surface: where = normal vector pointing outward

18. Dielectric-conductor boundary • Based on the figure in previous slide: • In a perfect conductor, • Hence, • This requires the tangential and normal components of E2 and D2 to be zero.

20. Capacitance • Capacitor – two conducting bodies separated by a dielectric medium ρs = Q / A ε1 ρs = surface charge density Q = charge (+ve / -ve) A = surface Area E = ρs/ε E E E ε2 E=0/V=0 on the surface

21. Capacitance • Capacitance is defined as: where: V = potential difference (V) Q = charge (C) C = capacitance (F)

22. Example 7 Obtain an expression for the capacitance Cof a parallel-plate capacitor comprised of two parallel plates each of surface area Aand separated by a distance d. The capacitor is filled with a dielectric material with permittivity ε.

23. Solution to Example 7 • expression for the capacitance C = Q/V and •  the voltage difference is • Hence, the capacitance is: ρs = Q / A

24. Example 8 Use image theory to determine E at an arbitrary point P (x, y, z)in the region z > 0 due to a charge Q in free space at a distance d above a grounded conducting plane.

25. Solution to Example 8 • Charge Q is at (0, 0, d) and its image −Q is at (0,0,−d) in Cartesian coordinates. Using Coulomb’s law, E at point P(x,y,z) due to two point charges:

26. Electrostatic potential energy • Assume a capacitor with plates of good conductors – zero resistance, • Dielectric between two conductors has negligible conductivity, σ≈ 0 – no current can flow through dielectric • No ohmic losses occur anywhere in capacitor • When a source is connected to a capacitor, energy is stored in capacitor • Charging-up energy is stored in the form of electrostatic potential energy in the dielectric medium

27. Electrostatic potential energy • Electrostatic potential energy, • The capacitance: • Hence, We for a parallel plate capacitor:

28. Image Method • Image theory states that a charge Q above a grounded perfectly conducting plane is equal to Q and its image–Q with ground plane removed.