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Electrostatics EEL 3472 Electrostatics Electrostatics An electrostatic field is produced by a static (or time-invariant) charge distribution. A field produced by a thunderstorm cloud can be viewed as an electrostatic field. Coulomb’s law deals with the force a point charge exerts on another point charge. The force F between two charges Q1 and Q2 is 1. Along the line joining Q1 and Q2 2. Directly proportional to the product Q1 Q2 of the charges 3. Inversely proportional to the square of the distance r between the charges F 2 kQ1Q2 r2 EEL 3472 Electrostatics In the international system of units (SI), Q1 and Q2are in coulombs (1C is approximately equivalent to 6x101 8 electrons since one electron charge e = 1.6019 x10 1 9 C), the distance r is in meters, and the force F is in newtons so that: k 1 4o o 8.854 x10 or k F 3 1 4o 1 2 10 9 F/m 36 the permittivity of free space 9x109 m/F Q1Q2 4o r 2 EEL 3472 Electrostatics Electric field intensity (or electric field strength) is defined as the force per unit charge that a very small stationary test charge experiences when it is placed in the electric field. F N V E lim , Qt 0 Q C m t In practice, the test charge should be small enough not to disturb the field distribution of the source. For a single point charge Q located at the origin E F Q er Qt 4or 2 where er is a unit vector pointing in the radial direction (away from the charge). o is valid for vacuum and air. For other materials o should be replaced by r o (typically r 1 100 ). 4 EEL 3472 Electrostatics Suppose that there are several charges Qi located at positions ri with respect to the origin. The observation point is at position r . The vector from the ith charge to the observation point is ( r ri ), and the distance is r ri . Source point The total field due to N charges is r ri E N i 1 Observation (field) point e rri 5 r ri r ri N i 1 Qi 4 r ri Qi 4 r ri 3 2 erri (r r i ) EEL 3472 Electrostatics Example: Find the electric field at P due to Q and –Q. The vector from Q to P lies in the x-y plane. Therefore, Ez (the z component of E ) vanishes at P. y components due to Q and –Q are equal and opposite and therefore add to zero. Thus, E has only an x component. E Q x E e x E cos 6 Q 4 (b2 d 2 ) Qb 3 4(b2 d 2 )2 1 2 2 b 2 1 2 2 (b d ) ExQ ExQ Ex ExQ ExQ 2ExQ 2Qb 3 4(b2 d 2 )2 EEL 3472 Electrostatics Various charge distributions. Charge density Consider a charged wire. Charge on an elemental wire segment l is q so that the charge per unit length q / l . The total electric field is the sum of the field contributions from the individual segments. E(r) i q 4 r ri 1 4 i If we let l 0 we get E(r) 7 1 4 3 (r ri ) (q / l )l r ri 3 (r ri ) (r)dl (r r) 3 r r q / l Position of the field point Position of the source point EEL 3472 Electrostatics If the charge is distributed on a surface (surface charge density q / A ), we can write E(r) 1 4 ( r)dA 3 r r (r r) 1 4 (r)dA 2 r r e rr Surface integral Unit vector directed from the source point to the field point If charge is distributed through a volume ( q / v) , E(r) 1 4 ( r)dv 3 r r (r r) 1 4 (r)dv 2 r r e rr Integral over the volume containing the charge 8 EEL 3472 Electrostatics Example: A circular disk of radius b with uniform surface charge density , lies in the x-y plane with its center at the origin. Find the electric field at P (z=h). z All the contributions in the x and y directions are cancelled, and the field at P contains only z component. Ez 1 4 r r b 0 h 2 u 2 9 h u 2 cos 2 2udu dA h 1 2 h2 u2 b 2 h b 1 2 0 h udu 2 0 h2 u2 h 1 1 h 1 2 h2 b2 h 2 h2 b2 3 2 1 2 EEL 3472 Electrostatics The Source Equation As we have seen, the electric field can be expressed as an integral of the charges that produce it. The charge can be expressed in terms of derivatives of the electric field: E source equation divergence operator Let E(x, y, z) E x (x, y, z)e x Ey (x, y, z)ey E z (x, y, z)e z Then 10 E E x Ey E z x y z scalar EEL 3472 Electrostatics The Source Equation continued Example: Let E be the field of a single point charge Q at the origin. Show that E is zero everywhere except at the origin. E Q 4 r 2 er Q 4 r 3 r er r r r xe x y ey z e z where E The x component of E is Ex Q x 3 4 (x 2 y 2 z 2 )2 E x Q y 2 z 2 2x 2 5 x 4 (x 2 y 2 z 2 ) 2 Ey Q x 2 z 2 2y 2 5 y 4 (x 2 y 2 z 2 )2 2 2 2 E z Q x y 2z 5 z 4 (x 2 y 2 z 2 )2 11 Q xe x y ey z e z 3 4 (x 2 y 2 z 2 )2 The divergence of a vector field at a given point is a measure of how much the field diverges or emanates from that point. E x Ey E z 0 x y z except at the origin (x=0,y=0,z=0) where the derivatives are undefined. EEL 3472 Electrostatics The Source Equation continued In cylindrical coordinates E 1 1 E EZ (EP ) z In spherical coordinates E 1 2 1 1 E ( r E ) (sin E ) r r sin r sin r 2 r Example: Let E be the electrostatic field of a point charge Q at the origin. Express this field in spherical coordinates and find its divergence. E Er Q 4r 2 E 0 Q er 2 4r E 0 E 1 Q 0 r 2 r 4 everywhere except at the origin, where E is undefined (because the field is infinite). 12 EEL 3472 Electrostatics Gauss’ Law Essentially, it states that the net electric flux through any closed surface is equal to the total charge enclosed by that surface. Gauss’ law provides an easy means of finding E for symmetrical charge distributions such as a point charge, an infinite line charge, and infinite cylindrical surface charge, and a spherical distribution of charge. Let us choose an arbitrary closed surface S. If a vector field V is present, we can construct the surface integral of V over that surface: V ds S The total outward flux of V through S The divergence of V (a scalar function of position) everywhere inside the surface S is V.We can integrate this scalar over the volume enclosed by the surface S. Vdv V 13 Over the volume enclosed by S EEL 3472 Electrostatics Gauss’ Law continued According to the divergence theorem V ds Vdv S V We now apply the divergence theorem to the source equation E integrated over the volume v: V E ds S Gaussian surface E dv Qenc V D E dv E ds S V E cos dA dv Qenc dA dS Gauss’ Law the angle between E and the outward normal to the surface S En - the normal component of E 14 EEL 3472 Electrostatics Gauss’ Law continued Example: Find the electric field at a distance r from a point charge q using Gauss’ law. Let the surface S (Gaussian surface) be a sphere of radius r centered on the o charge. Then 0 and cos 1 . Also E Er (only radial component is present). E cos dA S S Er dA Qenc q Since Er = constant everywhere on the surface, we can write Er dA S q Er q 4r 2 4r 2 15 EEL 3472 Electrostatics Gauss’ Law continued Example: A uniform sphere of charge with charge density o and radius b is centered at the origin. Find the electric field at a distance r from the origin for r>b and r<b. As in previous example the electric field is radial and spherically symmetric. (1) r > b 4b 3 4r E r o 3 2 2 r 0 0 2 2 sindd b o Er 3r 2 3 b r 0 2 sindrd d 0 r 0 decreases as 1 / r 2 (2) r < b Er ro 3 increases linearly with r Gauss’ law can be used for finding E when E is constant on the Gaussian surface. 16 EEL 3472 Electrostatics Gauss’ Law continued Example: Determine the electric field intensity of an infinite planar charge with a uniform surface charge density . coincides with the xy-plane The E field due to a charged sheet of an infinite extent is normal to the sheet. We choose as the Gaussian surface a rectangular box with top and bottom faces of an arbitrary area A, equidistant from the planar charge. The sides of the box are perpendicular to the charged sheet. E dS 17 Qenc EEL 3472 Electrostatics Gauss’ Law continued On the top face E dS E z e z dse z E z ds E dS E z e z dse z E z ds On the bottom face There is no contribution from the side faces. E dS 2E ds 2E z S The total enclosed charge is A 2E z A 18 z A S Qencl A Ez 2 E z e z ,z 0 E E z e z ,z 0 EEL 3472 Electrostatics Ohm’s Law Ohm’s law used in circuit theory, V=IR, is called the “macroscopic” form of Ohm’s law. Here we consider “microcopic” (or point form). (Applicable to conduction current only). Let us define the current density vector J . Its direction is the direction in which, on the average, charge is moving. J is current per unit area perpendicular to the direction of J . The total current through a surface is I J dS (if J J o , S a2, 0 ,I Ioa2 ) S According to Ohm’s law, current density has the same direction as electric field, and its magnitude is proportional to that of the electric field: J E E where E is the electric conductivity. The unit for E is siemens per meter ( S / m ( m)1 ). Copper Seawater Glass 19 5.80 x107 S/m 4 S/m 10 1 2 S/m EEL 3472 Electrostatics Ohm’s Law continued The reciprocal of E is called resistivity, in ohm-meters ( m). Suppose voltage V is applied to a conductor of length l and uniform cross section S. Within the conducting material J E E ( E constant) E J V l I V EE E S l l l I I RI V S ES “macroscopic” form of Ohm’s law Resistance of the conductor 20 EEL 3472 Electrostatics Electrostatic Energy and Potential A point charge q placed in an electric field E experiences a force: F qE If a particle with charge q moves through an electric field from point P1 to point P2, the work done on the particle by the electric field is W P2 P2 P1 P1 F dl q E dl This work represents the difference in electric potential energy of charge q between point P1 and point P2 W = W1 - W2 (W is negative if the work is done by an external agent) The electric potential energy per unit charge is defined as electric potential V (J/C or volts) P2 Electrostatic potential V1 V2 E dl difference or voltage P1 between P1 and P2 W1 W2 (independent of the path q q between P1 and P2 ) 21 EEL 3472 Electrostatics Electrostatic Energy and Potential continued V1 and V2 are the potentials (or absolute potentials) at P1 and P2, respectively, defined as the potential difference between each point and chosen point at which the potential is zero (similar to measuring altitude with respect to sea level). In most cases the zero–potential point is taken infinity. Example: A point charge q is located at the origin. Find the potential difference between P1 (a,0,0) and P2 (0,b,0) E due to q is radial. Let us choose a path of integration composed of two segments. Segment I (circular line with radius a) and Segment II (vertical straight line). V1 V2 P2 E dl P1 22 EEL 3472 Electrostatics Electrostatic Energy and Potential continued Example continued: Segment I contributes nothing to the integral because E is perpendicular to dl everywhere along it. As a result, the potential is constant everywhere on Segment I; it is said to be equipotential. On Segment II, e r e y , E dl dye y , and r y q q e ey r 4r 2 4y 2 qdy qdy E dl e e y y 4y 2 4y 2 b V1 V 2 qdy q 1 1 2 a 4y 4 a b dy 1 2 y y If we move P2 to infinity and set V2=0, V1 q 4a point where a is the distance between the observation point and the source In general, q 4 r r Vr qi i i 23 dl i for line charge ds for surface charge dv for volume charge EEL 3472 Electrostatics Gradient of a Scalar Field By definition, the gradient of a scalar field V is a vector that represents both the magnitude and direction of the maximum space rate of increase of V. The gradient operator () is a differential operator. It will act on the potential (a scalar) to produce the electric field (a vector). Let us consider two points P1(x1, y1, z1) and P2(x2 x, y2 y, z2 z) which are quite close together. If we move from P1 to P2 the potential will change from V1 to scalar V V V V x y z x y z V1 V where due to the change in z The displacement vector (express the change in position as we move from P1 and P2) is vector l x e x y e y z e z 24 EEL 3472 Electrostatics Gradient of a Scalar Field continued The relation between V and l can be expressed using the gradient of V: V V V V ex ey ez x y z vector V V l On the other hand, if we let P1 and P2 to be so close together that E is nearly constant, V V2 V1 (V1 V2 ) E l (potential difference equation) V V 2 1 E dl P 1 P2 E V The negative sign shows that the direction of E is opposite to the direction in which V increases. 25 EEL 3472 Electrostatics Gradient of a Scalar Field continued Example: Determine the electric field of a point charge q located at the origin, by first finding its electric scalar potential. V q 4 x 2 y 2 z 2 q 4r V qx x 4 x 2 y 2 z 2 V qy y 4 x 2 y 2 z 2 V qz z 4 x 2 y 2 z 2 V 3 2 3 2 3 2 V V V ex ey ez x y z E V E 26 q xe x y ey z e z q q r er 3 3 2 4 x 2 y 2 z 2 2 4 r 4 r EEL 3472 Gradient of a Scalar Field continued Example continued: Determine the electric field of a point charge q located at the origin, by first finding its electric scalar potential. dV V dl V dl cos vector the angle between dl and V dV V cos dl dV dV V dl dl MAX 0o 90 27 o dV 0 dl (V=const) V points in the direction of the maximum rate of change in V. V at any point is perpendicular to the constant V surface which passes through that point (equipotential surface). EEL 3472 Electrostatics Gradient of a Scalar Field continued The electric field is directed from the conductor at higher potential toward one at lower potential. Since E V , E is always perpendicular to the equipotentials. If we assume that for an ideal metal E , no electric field can exist inside the metal (otherwise J E E , which is impossible). Considering two points P1 and P2 located inside a metal or on its surface – and P recalling that V1 V2 E dl we find that V1=V2. This means the entire metal is 2 P1 at the same potential, and its surface is an equipotential. V 1 V V e e ez z V 1 V 1 V V er e e r r r sin For cylindrical coordinates V For spherical coordinates 28 EEL 3472 Electrostatics Capacitors Any two conductors carrying equal but opposite charges form a capacitor. The capacitance of a capacitor depends on its geometry and on the permittivity of the medium. C The capacitance C of the capacitor is defined as the ratio of the magnitude of Q of the charge on one of the plates to the potential difference V between them. Area A Q V E dS Qenclosed E dS E A A / S d d is very small compared with the lateral dimensions of the capacitor (fringing of E at the edges of the plates is neglected) S E A C d D A parallel-plate capacitor. Cylindrical Top face (inside the metal) Side surface dS Bottom face A E 0 Gaussian Surface E Close-up view of the upper plate of the parallel-plate capacitor. 29 EEL 3472 Electrostatics Laplace’s and Poisson’s Equations The usual approach to electrostatic problems is to begin with calculating the potential. When the potential has to be found, it is easy to find the field at any point by taking the gradient of the potential E V the divergence operator E vector the source equation V V the gradient operator the Laplacian operator 2V scalar If no charge is present 30 Poisson’s equation 0 , 2V 0 Laplace’s equation EEL 3472 Electrostatics Laplace’s and Poisson’s Equations continued The form of the Laplacian operator can be found by combining the gradient and divergence operators. V V V 2V V ex ey e z y z x 2V 2V 2V x 2 y 2 z 2 1 V 2 - in rectangular coordinates V 1 2V 2V 2 2 2 - in cylindrical coordinates z 1 2 V 1 V 1 2V -in spherical V 2 r sin r 2 sin2 2 r r r r 2 sin coordinates 2 Electrostatic problems where only charge and potential at some boundaries are known and it is desired to find E and V throughout the region are called boundary-value problems. They can be solved using experimental, analytical, or numerical methods. 31 EEL 3472 Electrostatics Method of Images A given charge configuration above an infinite grounded perfectly conducting plane may be replaced by the charge configuration itself, its image, and an equipotential surface in place of the conducting plane. The method can be used to determine the fields only in the region where the image charges are not located. Original problem Construction for solving original problem by the method of images. Field lines above the z=0 plane are the same in both cases. 32 By symmetry, the potential in this plane is zero, which satisfies the boundary conditions of the original problem. EEL 3472