(3-3) = (Gauss s law) (3-6)

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1 tatic Electric Fiels Electrostatics is the stuy of the effects of electric charges at rest, an the static electric fiels, which are cause by stationary electric charges. In the euctive approach, few funamental relations for an iealize moel are postulate as axioms, from which particular laws an theorems can be erive. Then the valiity of the moel an the axioms are verifie by the experiments. The steps involve in builing a theory base on an iealize moel are as follows:. Define some basic quantities. (E, q). pecify the rules of operations. (Vector analysis). Postulate some funamental relations. (Divergence equation, Curl equation) - Electrostatics in Free pace Here, electric fiel in free space is consiere. Note that the permittivity of the free space, enote by ε, is equal to (/6π) (F/m). First, electric fiel intensity is efine as the force per unit charge that a very small stationary test charge experiences when it is place in a region where an electric fiel exists. That is, F E lim (V/m) (-) q q Thus, E is proportional to an in the irection of the force F. Notice that the unit Newton/Coulomb V/m. An inverse relation of (-) gives F qe (N) (-) The two funamental postulates of electrostatics in free space specify the ivergence an the curl of E. They are ρ v ε E (-) an E, (-4) where ρ v enotes the volume charge ensity with the unit (C/m ). The efinition of ρ v is given by q ρ v lim (C/m ) v v (-4) asserts that static electric fiels are irrotational whereas (-) implies that a static electric fiel is not solenoial. These two equations are point relations or in ifferential forms. Taking the volume integral of both sies of (-) over a volume V yiels ρv Q E v v V V ε ε where Q is the total charge containe in V. Applying the ivergence theorem, one obtains Q E s (Gauss s law) (-6) ε which is a form of Gauss s law. Likewise, taking the surface integral of both sies of (-4) an applying tokes theorem yiels C E l, (-7) Permittivity is a physical quantity that escribes how an electric fiel affects, an is affecte by, a ielectric meium, an is etermine by the ability of a material to polarize in response to the fiel, an thereby reuce the total electric fiel insie the material. Thus, permittivity relates to a material's ability to transmit (or "permit") an electric fiel.

2 which asserts that the scalar line integral of the static electric fiel intensity aroun any close path vanishes. (-6), (-7) are of integral forms. ince the scalar prouct E l integrate over any path is the voltage along that path, i.e., V E l (V), C thus (-7) is equivalent to Kirchhoff s voltage law, i.e., the algebraic sum of voltage rops aroun any close circuit is zero. - Coulomb s Law Consier a single point charge q at rest in bounless free space. In orer to fin the electric fiel intensity ue to q, a spherical surface of an arbitrary raius r centere at q a hypothetical enclose surface (a Gaussian surface) aroun the source is rawn, upon which Gauss s law is applie to etermine the fiel. ince a point charge has no preferre irections, its electric fiel must be everywhere raial an has the same intensity at all points on the spherical surface. Applying (-6) to Fig. (a) yiels q q E s (ˆE ) ˆ r s r r or Er 4π ε s Er r. ε Therefore, q E rˆ E r rˆ (V/m) (-8) 4 πε r Fig. From (-8), the electric fiel intensity of a point charge is in the outwar raial irection an has a magnitue proportional to the charge an inversely proportional to the square of the istance from the charge. If the charge q is not locate at the origin, referring to Fig. (b), one obtains the electric fiel intensity at point P to be q E P aˆ qp r r'. r r' But a ˆqP r r' q( r r') EP 4 πε r r' (V/m) (-) Example - Determine the electric fiel intensity at P(-.,,-.) ue to a point charge of +5 (nc) at Q(.,.,-.5) in air. All imensions are in meters.

3 When a point charge q is place in the electric fiel of another point charge q, a force F is experience by q ue to E of q at q, which is given by qq F qe ˆ a (N) ; r r (-) (-) is a mathematical form of Coulomb s law : the force between two point charges is proportional to the prouct of the charges an inversely proportional to the square of the istance of separation. Example - The electrostatic eflection system of a cathoe-ray oscillograph is epicte in the right figure. Electrons from a heate cathoe are given an initial velocity u zˆu by a positively charge anoe. The electrons enter at z into a region of eflection plates where a uniform electric fiel E yˆ E is maintaine over a with w. Ignoring gravitational effects, fin the vertical eflection of the electrons on the fluorescent screen at zl. -. Electric fiel ue to a system of iscrete charges uppose an electrostatic fiel is create by a group of n iscrete point charges, q, q,, q n, locate at ifferent positions, the principle of superposition can be applie to fin the total electric fiel ue to this system of iscrete charges, which is given by q( r r ) q( r r ) q( r r n) E + + L+ or r r r r r r E n q k ( r r k k r rk ) (V/m) (-4) -. Electric fiel ue to a continuous istribution charges The electric fiel cause by a continuous istribution of charge as shown in the figure on the right can be obtaine by integrating the contribution of an element of charge over the charge istribution. Let ρ v be the volume charge ensity (C/m ), then the electric fiel intensity ue to qv at P is given by ρvv' E aˆ Therefore, n ρvv' E E aˆ ' ' ρ ' ' vv (V/m) ; V V V aˆ (-6)

4 For the charge istribute on a surface with a surface charge ensity ρ s (C/m ) (-6) becomes ss' E aˆ ρ (V/m) (-7) ' For a line charge with a line charge ensity ρ l (C/m), (-6) becomes ρ l' E aˆ l (V/m) (-8) ' L Example - Determine the electric fiel of an infinitely long, straight, line charge of uniform ensity ρ l (C/m) in air -4 Gauss s Law an Applications Gauss s law follows irectly from (-) an is given by Q E s (-6) ε Gauss s law asserts that the total outwar flux of the E-fiel over any close surface in free space is equal to the total charge enclose in the surface ivie by ε. The surface can be hypothetical close surface chosen for convenience, not necessarily be a physical surface. Gauss s law is useful in etermining E when the normal component of the electric fiel intensity is constant over an enclose surface. The first step to apply Gauss s law is to choose such surface, referre to as a Gaussian surface, an then evaluate both sies of (-6) in orer to etermine E. Example -4 Use Gauss s law for Example - 4

5 Example -5 Determine the electric fiel intensity ue to an infinite planar charge with a uniform surface charge ensity ρ s. Example -6 Determine the E fiel ue to a spherical clou of electrons with a volume charge ensity ρ insie an outsie. -5 Electric Potential ince ( V ) an E in electrostatics, one can efine a scalar electric potential V from (-4) such that E V (-6) Electric potential is relate to the work in carrying a charge from one point to another. ince the electric fiel intensity is the force acting on a unit test charge, the work require to move a unit charge from point P to P is given by W P E l (J/C or V) (-7) q P ince the static electric fiel is conservative, the line integral on the right oes not epen on the integration path, for instance integrations along path an path give the same result. Analogous to the concept of potential energy in mechanics, (-7) represents the ifference in electric potential energy of a unit charge between point P an point P. Let V enote the electric potential energy per unit charge, the electric potential, then V V P E l (V) (-8) P P since E l V aˆ l V V V. Thus, a potential ifference (electrostatic P P P l P P voltage) is equivalent to the electric potential energy per unit charge. Note that point P here is the reference zero-potential point. In most cases, the reference point is taken at infinity; this convention normally applies when the reference point is not specifie explicitly. Observations regaring electric potential. Because of the negative sign, the irection of E is opposite to the irection of increasing V. 5

6 . The irection of V is normal to surfaces of constant V, thus E is perpenicular to equipotential lines or equipotential surfaces. -5. Electric Potential ue to a charge istribution Let infinity be the reference point, then the electric potential of a point at a istance from a point charge q is given by q q V r ˆ r ˆr (V) r (-9) The potential ifference between points, P an P, at istances an, respectively, is given by q V VP V P (-) The electric potential ue to a system of n iscrete charges, q,, q n, is given by n q V k k r rk (V) (-) For continuous charge istributions in confine regions, electric potentials are given by ρvv' V 4 (V/m) (volume charge) πε V ' (-8) ρss' V ' (V) (surface charge) (-9) ρ ' V 4 l l πε L' (V) (line charge) (-4) Example -7 [Electric ipole moment] Electric potential ue to an electric ipole consisting of charges +q an q with a small separation of (assume >> ) Example -8 Obtain a formula for the electric fiel intensity on the axis of a circular isk of raius b that carries a uniform surface charge ensity ρ s. 6

7 -6 Material Meia in tatic Electric Fiel Consier energy ban theory of solis base on soli state physics as shown in Fig., electrical materials can be classifie into types, namely, conuctors, ielectrics (or insulators), an semiconuctors. Figure : Energy ban structure -6. Conuctors in tatic Electric Fiel Assume that some electric charges are introuce in the interior of a goo conuctor. An electric fiel will be set up an create a force that causes the movement of charges. This movement will continue until all charges reach the conuctor surface an reistribute in such a way that both the charge an the fiel insie vanish. Hence, ρ ;E v When there are no free charges in the interior of a conuctor ( ρ v ), E must be zero accoring to Gauss s law. Furthermore, uner static conitions the E fiel on a conuctor surface is everywhere normal to the surface, otherwise there exists a tangential force that moves the charges. Consier the bounary conitions at the interface between a conuctor an free space as shown in Fig.. Integrating E along the contour abca an taking the limit as h yiel Figure : A conuctor-free space interface lim h abca E l E w or t Et Which says that the tangential component of the E fiel on a conuctor surface is zero uner static conitions. In other wors, the surface of a conuctor is an equipotential surface. Next, integrating E on the Gaussian surface in the figure an taking the limit as h : ρ s ρ s E s En or E n ε ε A ielectric is a nonconucting substance, i.e. an insulator. The term was coine by William Whewell in response to a request from Michael Faraay. Although "ielectric" an "insulator" are generally consiere synonymous, the term "ielectric" is more often use to escribe materials where the ielectric polarization is important, such as the insulating material between the metallic plates of a capacitor, while "insulator" is more often use when the material is being use to prevent a current flow across it. 7

8 Example -9 A positive point charge Q is at the center of a spherical conucting shell of an inner raius i an an outer raius o. Determine E an V as functions of the raial istance r. Figure : Example Dielectrics in tatic Electric Fiel All material meia are compose of atoms with a positively charge nucleus surroune by negatively charge electrons. In the absence of an external electric fiel, the molecules of ielectrics are macroscopically neutral. The presence of an electric fiel causes a force on each charge particle an results in small isplacements of positive an negative charges in opposite irections. These are boun charges. The isplacements polarize a ielectric material an create electric ipoles (i.e., polarization). The molecules of some ielectrics possess permanent ipole moments, even in the absence of an external electric fiel. uch molecules are calle polar molecules, in contrast to nonpolar molecules. An example is the water molecule H O. Generally, ielectric materials consist of both polar an nonpolar molecules (Fig. 5). When there is no external fiel, ipoles in polar ielectrics are ranomly oriente (Fig. 6 (a)), proucing no net ipole moment macroscopically. An applie electric fiel will ten to align the ipoles with the fiel as shown in Fig. 6 (b). proucing the nonzero net ipole moment (Fig. 7). Fig. 5: Molecules in ielectrics 8

9 Figure 6: Polar molecule Figure 7: Interior of a ielectric meium A polarization vector P is efine as N v p k k v v P lim (C/m ) where N is the number of molecules per unit volume an the numerator represents the vectopr sum of the inuce ipole moments containe in a very small volume v. The vector P is the volume ensity of electric ipole moment. The ipole moment p prouces an electric potential P rˆ V v' Thus, the potential ue to the polarize ielectric is given by P rˆ V v' ' V Interpretation of the effects of the inuce electric ipoles:. Equivalent polarization surface charge ensity ρ P ˆ (C/m ) ps a n. Equivalent polarization volume charge ensity ince Q P aˆ ( ) ns P v ρ pvv, V V Thus, one can efine the polarization volume charge ensity as ρ P (C/m ) pv It follows that total charge ρ + ˆ pss ρ pvv P a ns ( P) v, V V i.e., the total free charge of the ielectric boy after polarization must remain zero. Example - The polarization vector in a ielectric sphere of raius is P xˆp. Determine a) the equivalent polarization surface an volume charge ensities an b) the total equivalent charge on the surface an insie of the sphere 9

10 -7 Electric Flux Density an Dielectric Constant In ielectrics, E ( ρ v + ρ pv ), ε but since ρ pv P, ( ε E + P) ρv. Here, one can efine a new funamental fiel quantity, Electric Flux Density (or electric isplacement) D to be D ε E + P (C/m ) It follows that D ρ v Applying the ivergence theorem yiels D v D s ρ v Q. V Hence, V D s Q (C) v In linear an isotropic meia, P can be given in terms of E as P ε E χ e where χ e is calle electric susceptibility (imensionless). Here, D can be rewritten as D ε ( + χ ) E ε ε E εe (C/m ) e r where ε + χ r e ε ε is calle relative permittivity or ielectric constant (imensionless). In general, ielectric materials can be classifie base on the property of ielectric constants into Linear : ielectric constant oesn t change with applie electric fiel non-linear Isotropic: ielectric constant oesn t change with irection anisotropic Homogeneous: ielectric constant oesn t change from point to point inhomogeneous -7. Dielectric trength If the electric fiel is very strong, it will pull electrons completely out of molecules. The electrons will accelerate uner the influence of the electric fiel, collie violently with the molecular structure an avalanche effect of ionization ue to collisions may occur. The material will become conucting an large currents may result; this phenomenon is calle a ielectric breakown. The maximum electric fiel intensity that a ielectric material can withstan without breakown is the ielectric strength. For instance, the ielectric strength of air at the atmospheric pressure is (kv/mm). Example - Consier two spherical conuctors with raii b an b (b > b ) that are connecte by a conucting wire. The istance of separation between the conuctors is assume to be very large in comparison to b so that the charges on the spherical conuctors may be consiere as uniformly istribute. a) the charges on the two spheres, an b) the electric fiel intensities at the sphere surfaces.

11 Example -9* A positive point charge Q is at the center of a spherical ielectric shell, with a ielectric constant of ε r an (inner, outer) raii, ( i, o ), respectively. Determine E, V, D, P as functions of the raial istance r. Figure 8 : Example -9* -8 Bounary Conitions for Electrostatic Fiels Consier the bounary conitions at the interface between two ielectric meia as shown in Fig. 9. Integrating E along the contour abca an taking the limit as h yiel lim E l E w + E ( w) or E E h t t abca which says that the tangential component of the E

12 fiel is continuous across the interface. If ε, ε enote the permittivities of meia,, respectively, then D t D t ε ε Next, integrating E on the Gaussian surface in the figure an taking the limit as h : D s D aˆ + D aˆ ( n n) aˆ ( D D ) ρ n Thus, ˆ n ( D D ) ρ s or D n Dn ρs a (C/m ) s Figure 9 : An interface between two meia where ρ s enotes the surface charge ensity on the interface. Example - A lucite sheet (ε r.) is introuce perpenicularly in a uniform electric fiel E o xˆe in free space. Determine E i,d i,p i insie the lucite. Example -4 Two ielectric meia with permittivities ε an ε are separate by a charge free bounary. The electric fiel intensity in meium at the point P has a magnitue E an makes an angle α with the normal. Determine the magnitue an irection of E at point P in meium.

13 -9 CAPACITANCE AND CAPACITO It is known from -6 that a conuctor in a static electric fiel is an equipotential boy an that charges on a conuctor will istribute themselves in such a way that the electric fiel insie vanishes. uppose the potential ue to a charge Q is V, then increasing the total charge by a factor k only increases the surface charge ensity without changing the charge istribution. It is also note that increasing Q also leas to increasing E an thus V also increases. eciprocally, increasing V by a factor of k leas to increase in Q. ρs kρs V l n kv ke l n, E an ε Thus, one can conclue that the Q/V ratio remains unchange. This ratio is calle the capacitance of the isolate conucting boy, which has the unit Fara (F), or C/V. Using C, one can write Q CV Of consierable importance in practice is the Capacitor (or Conenser) as shown in Fig.. Figure : A two-conuctor capacitor ε Here, the capacitor consists of two conuctors separate by free space or a ielectric meium. When a c voltage source is applie between conuctors, a charge transfer occurs, resulting in +Q on one conuctor an Q on the other. Note that the fiel lines are perpenicular to the conuctor surfaces. Let V be the potential ifference between two conuctors, then the capacitance C is given by Q C (F) The capacitance of a capacitor epens on the geometry an the permittivity of the meium. Example -5 A parallel-plate capacitor consists of two parallel conucting plates of areas separate by a uniform istance. The space between the plates is fille with a ielectric of a constant permittivity ε. Determine the capacitance. V Example -6 A cylinrical capacitor

14 Example -6* A spherical capacitor - ELECTOTATIC ENEGY AND FOCE ince electric potential at a point in an electric fiel is the work require to bring a unit charge from infinity (the reference point) to that point, to bring a charge Q from infinity against the fiel of a charge Q in free space to a istance requires the work of amount Q Q W QV Q Q QV, which is path-inepenent. The work is store in the assembly of two charges as potential energy, W ( QV + QV ) Here, if a charge Q is brought from infinity to a point that is from Q an from Q, then an aitional amount of work is require that equals Q Q W Q + V Q The potential energy store in charges is given by QQ QQ QQ W W + W πε Q Q ( QV + Q V + Q V ) Q + Q + Q Q + Q + Q Q + Note that V, the potential at the position of Q, is cause by charges Q, Q, an it is ifferent from the V in the two-charge case. Using the same proceure, the potential energy (electrostatic energy) of a group of N iscrete point charges can be given by N We QkVk (J) k Likewise, the potential energy ue to continuous charges can be given by We ' (J) ρvvv V ' ince the I unit for energy, Joule (J), is too large, a more convenient unit, electronvolt (ev), which is the energy or work require to move an electron against a potential ifference of one volt, i.e., 4

15 (ev).6 9 (J) is use instea. Example -7 Fin the energy require to assemble a uniform sphere of charge of raius b an volume charge ensity ρ v. -. Electrostatic Energy in Terms of Fiel Quantities ince D ρ v, We ( D ) Vv', V ' using ( VD) V D + D V yiels We ( VD ) v' D Vv' VD s' + D Ev'. V ' V ' ' V ' ince at least D is proportional to /r an V is proportional to /r, let V be the sphere of raius r, an taking r, the first term of the right han sie vanishes. Hence, We ' (J) D Ev V ' For a linear, isotropic meium, We ' (J) ε E v V ' Here, one can efine electrostatic energy ensity w e as we ε E (J/m ); We wev' V ' Example -8 A parallel-plate capacitor Example -9 A cylinrical capacitor (Figure ) -. Electrostatic Forces ecall that Coulomb s law governs the force between two point charges, but it might be har to etermine the force using Coulomb s law in a more complex system of charge boies. In such cases, the following principle is useful. Principle of Virtual Displacement : calculate the force on an object in a charge system from the electrostatic energy of the system 5

16 Consier an isolate system of charge conucting, as well as ielectric, boies separate from one another with no connection to the outsie worl. The mechanical work one by the system to isplace one of the boies by a ifferential istance l (a virtual isplacement) is given by W F l (-) Q where F Q enotes the total electric force. ince it is an isolate system with no external supply of energy, the mechanical work must be one at the expense of the store electrostatic energy, i.e., W W F l. (-) e Q Writing the force in terms of the graient of the work, one can write We ( We ) l (-4) ince l is arbitrary, comparison of (-) an (-4) yiels F (N) Q W e Example - the force on conucting plates of a parallel-plate capacitor - OLUTION OF ELECTOTATIC BOUNDAY-VALUE POBLEM o far, techniques for etermining E, D, V, etc for a given charge istribution have been iscusse. In many practical problems, the charge istribution is not known everywhere. In such cases, ifferential equations that govern the electric potential in an electrostatic situation are formulate, an the bounary conitions are applie to obtain what are calle bounary-value problems. -. Poisson s an Laplace s Equations In Electrostatics, D ρv ; E ; E V In linear, isotropic meium, since D εe, D ( ε E) ( ε V ) ρ v Hence, one obtains Poisson s equation an Laplace s equation (ρ v case) as follows: ρ v Poisson s : V ( ' ) ; ( Laplace' s) : V ε where (el square) is calle Laplacian operator. -.* Uniqueness Theorem Uniqueness theorem asserts that a solution of an electrostatic problem satisfying its bounary conitions (Poisson s equation or Laplace s equation) is the only possible solution, irrespective of the metho by which the solution is obtaine. Proof uppose a volume τ is boune outsie by a surface o which may be a surface at infinity. Insie the close surface o there are a number of charge conucting boies with surfaces,,, n at specifie potentials, as epicte in Fig.. 6

17 Figure proof of Uniqueness theorem ( V V ) V Now assume that, contrary to the uniqueness theorem, there are two solutions, V an V, to Poisson s equation in τ: ρv ρv V ; V ε ε Also assume that both V an V satisfy the same bounary conitions on,, n an o. Let V V - V, then V ; V in in, K, n τ Using ( f ) f A + A f V yiels A an letting fv, A ( V ) + ( V ) ( V ) V V + V V Integrating both sies of the equation above yiels τ ( V V ) v ( V V ) s τ V ince V on,,, n an on r ; V / r; V / r ; s r. v Thus, the integral on the left han sie vanishes. ince V is nonnegative, V must be ientically, which means V has the same value at all points in τ as it has on the bouning surfaces,,, n, where V. Thus, V everywhere, an therefore V V, i.e., only one solution exists. Example - parallel conucting plates separate by with ρ v - ρ y/ Example - Two infinite insulate conucting plates maintaine at potentials an V (Figure ) Fin the potential istribution for <φ <α an α <φ<π. Figure Example - Example - Given the inner an outer raii of two concentric, thin, conucting, spherical shells ( i, o ), 7

18 espectively, an the space between the shells is fille with a ielectric. Determine the potential istribution in the ielectric material by solving Laplace s equation. Example -6** A spherical capacitor -.5 Metho of Images The metho of images is the technique applie to bounary value problems by replacing bounary surfaces with appropriate image charges, instea of attempting to solve a Poisson s or Laplace s equation. Example -4 Point Charges Near Conucting Planes as shown in Fig. (a) (a) Physical arrangement (b) Image charge an fiel lines Fig. A formal proceure woul require the solution of Poisson s equation in the y > region with bounary conitions V at y an at infinity. Here, if an appropriate image charge can be use to replace the conucting plane such that all bounary conitions are satisfie, then the solution woul be obtaine in a straightforwar manner. uppose one replaces the conuctor with the charge Q at (,-,), then the potential at a point P(x, y, z) is given by Q V ( x, y, z) + Q. y x + ( y ) + z x + ( y + ) + z y < Note that the conition V at y is satisfie. Then, E for y is given by Q xˆ x + yˆ( y ) + zˆ z xˆ x + yˆ( y + ) + zˆ z E V / / x + ( y ) + z x + ( y + ) + z [ ] [ ] Q Hence, the surface charge ensity becomes ρ s ε Ey. y π x + + z / ( ) 8

19 Line charge near a parallel conucting cyliner Consier the problem of a line charge ρ l locate at a istance from the axis of a parallel, conucting, circular cyliner of raius a. Both are assume to be infinitely long. Fig. (a) shows a cross section of this arrangement. To apply the metho of images, first observe that () The image must be a parallel line charge insie the cyliner in orer to make the cylinrical surface at r a an equipotential surface. Let call this image line charge ρ i () Because of the symmetry with respect to the line OP, the image line charge must lie somewhere along OP, say at a point P i, which is a istance i from the axis (Fig. (b)). The unknowns neee to be etermine here are ρ i an i. First, let ρ ρ, then the potential at a istance r from a line charge of ensity ρ l is given by r ρ r l ρl V Err r r πε r r πε i l r ln r Thus, the potential at point M can be foun by aing contributions of ρ l an ρ i, i.e., ρl r ρl r ρl ri VM ln ln ln πε r πε ri πε r In orer for an equipotential surface to coincie with the surface ra, r i /r must be a constant. The point P i must be locate such that OMP i is similar to OPM, i.e., OMP i OPM. Hence, Pi M OPi OM ri i a ; or constant PM OM OP r a Therefore, i a /. The point P i is calle inverse point of P with respect to a circle of raius a. Example -5 Capacitance per unit length between two long parallel circular conucting wires of raius a 9

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