Dipoles and Dielectrics

Kenneth H. Carpenter Department of Electrical and Computer Engineering Kansas State University September 22, 2008

1 Dipoles

A dipole is a pair of opposite signed, equal magnitude charges, located near each other. Such a pair has the dipole moment given by

p = q d pˆ (1) where q is the magnitude of the charges, d is the distance separating them, andp ˆ is a unit vector pointing from the negative charge of the pair to the positive charge. z

q When q is at position R and −q is at R R + − + −q then R− R p = q(R+−R−) = qd, for d = (R+−R−) y (2) the vector distance between charges. In the limit |d| → 0 while q → ∞ so that |p| remains x finite, the result is the ideal point dipole.

1.1 The potential due to a single dipole When one wishes to find the potential V (R) due to dipole p, one adds the potentials due to the two charges. Taking the reference for potential to be

1 EECE557 Dipoles and dielectrics – supplement to text - Fall 2008 2 at infinity, one has q  1 1  V∞(R) = − 4π0 |R − R+| |R − R−|   q 1 1 = q − q  4π0 2 2 2 2 R − 2R · R+ + R+ R − 2R · R− + R−   "  2#−1/2 "  2#−1/2 q R+ R+ R− R− =  1 − 2Rˆ · + − 1 − 2Rˆ · +  4π0R R R R R ! q  R   R  R 2 = 1 + Rˆ · + − 1 + Rˆ · − + O d 4π0R R R R  2! q ˆ Rd = 2 R · (R+ − R−) + O , (3) 4π0R R ! R 2 where O d means terms on the order of (R /R)2 or (R /R)2 or R + − higher powers of them. When R >> R+ and R >> R− then the result reduces to p · Rˆ V∞(R) = 2 . (4) 4π0R

Note that in eq.(4) the requirement that R >> R+ and R >> R− means not only that the dipole becomes a point dipole, but the dipole’s location goes to the origin, provided R remains finite.

1.1.1 Potential of a single dipole at a general position z

When p is at position R then p p Rp p · (R − Rp) V∞(R) = . (5) 3 R y 4π0|R − Rp)| x EECE557 Dipoles and dielectrics – supplement to text - Fall 2008 3

1.2 Dipole density When dealing with a material having large numbers of dipoles, it is conve- nient to define a dipole moment density as a macroscopic quantity in the same manner as is done for charge density: P p P = lim i i . (6) ∆V →δ>0 ∆V One may now use the dipole moment density times a volume increment to represent the dipole term in eq.(5) to obtain an increment of potential to integrate to find the effect of the dipole distribution: ZZZ 1 P (Rp) · (R − Rp) V∞(R) = dVp, (7) 3 4π0 |R − Rp)| where the dipole moment density P is a function of the coordinates Rp over which the integration is performed.

2 Dielectric materials

Dielectric materials are ones that do not conduct (at least in a limiting sense) electric current, but contain charges that are in dipoles or can be formed into dipoles by an electric field. In general we represent the dipoles in the dielectric by the dipole moment distribution or density, P . Thus the contribution the dipoles in a material make to the electrostatic potential is given by eq.(7) with the volume of integration being the volume of the dielectric.

2.1 Alternate representation of dielectric dipole’s ef- fects Instead of using eq.(7) to find the effect of the dipoles of a dielectric on the fields, one usually transforms eq.(7) in the following way. First note the following “trick” with vector fields:

1 (R − Rp) ∇p = , (8) 3 |R − Rp| |R − Rp| where the gradient is with respect to the coordinates of Rp. EECE557 Dipoles and dielectrics – supplement to text - Fall 2008 4

Next note the identity

∇ · (bA) ≡ A · ∇b + b · ∇ · A. (9)

Now use eq.(8) in eq.(7) and then identify b as 1/|R − Rp| and A as P to obtain 1 ZZZ 1 V∞(R) = P (Rp) · ∇p dVp 4π0 |R − Rp| ZZZ ZZZ 1 P 1 ∇p · P = ∇p · dVp − dVp 4π0 |R − Rp| 4π0 |R − Rp| 1 ZZ P · dS 1 ZZZ −∇ · P V (R) = + p dV (10) ∞ 4π 4π p 0 h|R − Rp| 0 |R − Rp| where the divergence theorem has been used at the last step. Now if one identifies

ρsp = P · nˆ and ρvp = −∇ · P (11) as surface and volume charge densities, but densities due to the dipoles, these polarization charge densities may be used in calculating the electric fields’ contribution from the dipoles rather than using the dipole moment distribution directly.

3 Modified equations when dielectrics present

When a dielectric material is present, the relationship between electric flux density and its sources may be written as

∇ · E = ρv/0 = ρvc/0 + ρvp/0, (12) where the charge density has been split into that due to the charges in the dipoles and any other charges present. (These other charges are sometimes called “conduction charges,” hence the sub-c for them.) Now using eq.(11) in eq.(12) one has ∇ · (0E) = ρvc − ∇ · P or

∇ · (0E + P ) = ρvc. (13) Next, define D = 0E + P (14) EECE557 Dipoles and dielectrics – supplement to text - Fall 2008 5

and redefine ρv to be just that part of the charge density that is not due to charges in dipoles and one has

∇ · D = ρv. (15)

3.1 Relationship of dipole density to fields In a linear material, one may write

P = χe0E. (16) Then one may define

r = 1 + χe and  = r0 (17) to obtain D = E. (18) Gauss’s law, in terms of D ,becomes ZZ ZZZ D · dS = ρvdV = Qenclosed. (19) h One may solve for the fields in the presence of symmetric dielectric and charge distributions by using Gauss’s law in this form.

4 Problems

1. Given a dipole p = 3( nC m)ˆy is located at (1 m, 2 m, −3 m)

(a) find the expression for V∞(R). (b) find E as −∇(V ).

2. A material has r = 3. Find P at a point where E = 3( V/m)ˆx. 3. Charge q = 1 nC is at the origin. Region R < 0.1 m is vacuum. Region 0.1 m < R < 0.5 m has r = 2.5. Region 0.5 m < R is vacuum. (a) Find D everywhere. (b) Find E everywhere. (c) Find V at R = 2 m, with the reference at infinity.

(d) Find ρsp on the dielectric surfaces at R = 0.1 m and R = 0.5 m.