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

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 electric dipoleNote, an electric dipole has been created !

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, ε 0 ε = ε r ε 0

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 : electric current J(r )Moving charge is electric current J(r ).

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 Boundary condition Dielectric and dielectric Dielectric and conductor Conductor and conductor

Dielectric- dielectric boundary Interface between two dielectric media

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

ε2 ε2 z xy -plane E1E1 E2E2 E xy1 E z1 E z2 E xy2 ε1 ε1

ε 2 = 5 ε 0 xy E1E1 E2E2 ε1=2ε0ε1=2ε0 Example 1: 1) Find E 2 in the dielectric, when E 1 = 3ax+4ay+5az, 2) And find Ө 1 and Ө 2. Solution: 1)E xy1= E xy2 thus,E xy2 = 3ax+4ay 2)E z1 = 5az, but, E z2 = ?? 2ε 0 (5az) = 5ε 0 (E z2 ) E z2 = 2az thus, E 2 =3az+4ay+2az Ө1Ө1 Ө2Ө2 z

Find E 1 if E 2 = 2x -3y +3z with  s = 3.54 x (C/m 2 ) And find Ө 1 and Ө 2 ε 2 = 8 ε 0 xy E1E1 E2E2 ε1=2ε0ε1=2ε0 Ө1Ө1 Ө2Ө2 z

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

xy z J z1 J xy1 J xy2 J z2 J1J1 J2J2

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.

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

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

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.

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

Dielectric-conductor boundary Based on the figure in previous slide: In a perfect conductor, Hence, This requires the tangential and normal components of E 2 and D 2 to be zero.

Capacitance Capacitor – two conducting bodies separated by a dielectric medium E=0/V=0 on the surface E E E ε1ε1 ε2ε2 E = ρs/ε ρs = Q / A ρs = Q / A ρs = surface charge density Q = charge (+ve / -ve) A = surface Area

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

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

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

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.

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:

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

Electrostatic potential energy Electrostatic potential energy, The capacitance: Hence, W e for a parallel plate capacitor:

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.