UNIT-1 Electrostatics Structure of the Unit 1.0 Objectives 1.1 Introduction 1.2 Electric Field 1.3 Gauss’s Law 1.4 Illustrative Examples 1.5 Self Learning Exercise-I 1.6 Scalar Potential 1.7 Electrostatic Boundary Conditions 1.8 Discontinuity in Potential due to Dipole Layer 1.9 Illustrative Examples 1.10 Self Learning Exercise-II 1.11 Summary 1.12 Glossary 1.13 Answers to Self Learning Exercises 1.14 Exercise 1.15 Answers to Exercise References and Suggested Readings 1.0Objectives1.0 Objectives This unit constitutes the basic concepts of static (time invariant )electric field and potential. One can learn the Gauss’s law and its applications. We learn the usefulness of spherical and cylindrical coordinates for solving the certain kinds of problems in electrostatics. The better physical insight of behaviour of electric field and potential across the interface can be got by studying the boundary conditions at the interface. 1 1.1Introduction1.1 Introduction This unit introduces Coulomb’s law ,Gauss’s Law and boundary conditions on electric fields and potentials. Gauss’s law is developed and shown in both integral and differential form. The concept of circulation of the electric field is related to its conservative nature is discussed. The concept of boundary conditions for electric field and potential is introduced in this unit. 1.21.2 Electric Electric Field Field According to Coulomb’s law , electrostatic force F between two point charges q and Q which are placed in free space at a distance r is expressed mathematically as Qq F k r 2 This force acts along the line joining the charges. 1 N m2 k 9 109 4 C 2 0 C 2 8.854 1012 0 N m2 The constant 0 is called permittivity of free space. Force on q2 due to q1 can be written in vector form as 1q q 1 q q F 1 2rˆ 1 2 r (1) 42 4 3 0r 0 r q .2 r 1 2 r 2 .q 1 r 1 O rig in Figure1.1 2 1q q 1 q q 1 2ˆ 1 2 F1 2 2r 1 2 3 r 1 2 40r 1 2 4 0 r 1 2 1 q1 q 2 F1 2 3 r 2 r 1 4 r r 0 2 1 where r1 2 r 2 r 1 =Position vector of q2 −Position vector of q1 r1 2 r 2 r 1 From eq.(1) taking q1 Q(Source charge) and q2 q (test charge),we have 1 qQ F rˆ 4 2 0 r 1 Q ˆ F q 2 r 4 0 r F qE (Force on the test charge q) 1 Q E rˆ (2) 4 2 0 r F and E q E is called electric field (or electric field intensity) due to point charge Q. N SI unit of E is . C Thus “Electric field E at a point is the force experienced per unit charge at rest state at that point of space.” E E() r has a value at each point in the space, so it is called vector point field. For definition of electric field , we can write F E lim q0 q The test charge q should be infinitesimally small because large value of the test charge will disturb the original charge distribution of primary charges that produces E . 3 If there are more than two charges, then we use principle of superposition for determination of the force on a particular charge. If there are N point charges QQQQ1, 2 , 3 ,... N (source charges) placed respectively at distances r1, r 2 , r 3 ,... rN from charge q then by principle of superposition total force on the charge q(test charge)is given by FFFFF ... 1 2 3 N qQ qQ qQ 1ˆ 2 ˆN ˆ F k2r1 k 2 r 2 .... k 2 r N r1 r 2 r N Here position vectors of QQQQ, , ,... are r, r , r ,... r respectively and test 1 2 3 N 1 2 3 N charge q is located at position r , then r r, r r ,.... r1 1 r 2 2 QQ Q F q k1ˆ k 2 ˆ .... k N ˆ 2r1 2 r 2 2 r N r1 r 2 r N F qE QQ Q 1ˆ 2 ˆN ˆ where E k2r1 k 2 r 2 .... k 2 r N r1 r 2 r N Q E k i ˆ 2 r i r i or EEEEE1 2 3 ... N Above expression represents the principle of superposition for electric field. For continuous charge distribution 1 dq E rˆ 4 2 0 r Now we consider the Coulomb’s law for the general case of volume charge. dq Volume charge density is (in C/m3) ,where differential charge dq is d present in a differential volume d The electric field at a point r x,, y z in terms of integral over the volume charge distribution x,, y z is written as 4 1 r ˆ E() r r d 2 (3) 4 0 r For given charge distribution, E() r is a function of unprimed coordinates x,, y z r is a function of primed coordinates x,, y z ˆ ˆ ˆ r rr xxi yyj zzk ’ Volume elementd dx dy dz For line charge distribution 1 r ˆ E() r r dl 2 4 0 L r dq Line charge density is (in C/m),where differential charge dq is present on dl a differential length dl . For surface charge distribution 1 r ˆ E() r r dS 2 4 0 S r dq Surface charge density is (in C/m2),where differential charge dq is present dS on a differential area dS . 1.31.3 Gauss’s Gauss’s Law Law The theorem is stated mathematically as follows – q E. d S enclosed (4) 0 where qenclosed = q inside = algebraic sum of charges inside the enclosed volume. “That net outward electric flux through a closed surface is equal to the sum of the charges inside the enclosed volume divided by the permittivity of free space”. Here such a hypothetical closed surface is known as Gaussian Surface. Gauss’s law is the easiest way of calculating electric field in situations in which charge distribution has symmetry such as spherical distribution of charge, an infinite line charge etc. Gauss’s law is also true in non symmetrical situations, but 5 in that case it will not be so useful for the evaluation of electric fields. Important Points: 1. Electric flux E. d S is independent of the size and shape of the Gaussian surface as long as q enclosed is same. 2. Electric flux does not depend on the location of charge inside the closed surface S, whereas electric field at each point on surface S, is dependent on location of charge. 3. Electric flux is unaltered by the charge outside the closed surface S, but outside charge contributes in electric field at each point on the surface S. The Concept of Solid Angle: Solid angle is analogous in three dimensional of the ordinary two dimensional angle. Now we consider an infinitesimal small area dS which subtends an infinitesimal small solid angle d at a point O (see Fig. 1.2). Figure1.2 Here r is the distance from the vertex O to the surface element dS. Solid angle d is related to opening of cone around its vertex. Mathematically, we have dS d r 2 Unit of solid angle is “Steradian” . From the definition of solid angle, it is obvious that the solid angle is dimensionless quantity. (i) Suppose an elemental area vector vector d S makes an angle with radial vector r (see figure 1.3) 6 Figure1.3 Projection area perpendicular to r is dS cos . Now solid angle d is the ratio of this projected area to r 2 dScos d S . r i.e. d (5) r2 r 2 (ii) In figure 1.4, area elements dS1, dS2 and dS3 subtend same solid angle at the point O, because for them, opening of cone around its vertex O is same. Figure1.4 (iii)Solid angle subtended by sphere at the centre is4 dS d 2 r 1 dS r = constant 2 r 1 (4r 2 ) 4 r 2 7 4 = complete (full) solid angle Whatever may be shape or size of a closed surface, above result holds at any internal point surrounded by the closed surface. Therefore we have general result: Entire closed surface subtends solid angle 4 at an internal point (6) 0 at an external point Proof of Gauss’s Theorem:- We consider a point charge q surrounded by a hypothetical closed surface S of an arbitrary shape as shown in Figure1.5 Figure1.5 Electric flux through an infinitesimal area element dS is d , then d E.d S EdS cos where is the angle between electric field E and area vector d S 1 q Due to the point charge q, electric field at a distance r is given by E 2 r 4 0 r 1 q d EdS cos dS cos (7) 4 r 2 0 8 Since electric field is directed along r , therefore is the also angle between rˆ and d S .Here area element d S subtends an infinitesimal solid angle d at the point charge q . d S .rˆ From eq.(5) ,we have d r 2 dS cos d (8) r 2 From eq.(7) & (8) 1 d qd (9) 4 0 Total flux through the closed surface S is 1 d qd S S 4 0 q d 4 0 q (10) 4 0 Here the hypothetical closed surface S subtends the total solid angle at the point of location of the charge q.