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- 1. ELECTROSTATICSGauss’s Law and ApplicationsThough Coulomb’s law is fundamental, one ﬁnds it cumbersome to use it to cal-culate electric ﬁeld due to a continuous charge distribution because the integralsinvolved can be quite difﬁcult. An alternative but completely equivalent formula-tion is Gauss’s Law which is very useful in situations which exhibit certain sym-metry.Electric Lines of Force :Electric lines of force (also known as ﬁeld lines) is a pictorial representation of theelectric ﬁeld. These consist of directed lines indicating the direction of electricﬁeld at various points in space.• There is no rule as to how many lines are to be shown. However, it iscustomary to draw number of lines proportional to the charge. Thus if Nnumber of lines are drawn from or into a charge Q, 2N number of lineswould be drawn for charge 2Q.• The electric ﬁeld at a point is directed along the tangent to the ﬁeld lines.A positive charge at this point will move along the tangent in a directionindicated by the arrow.• Lines are dense close to a source of the electric ﬁeld and become sparse asone moves away.• Lines originate from a positive charge and end either on a negative chargeor move to inﬁnity.1
- 2. • Lines of force due to a solitary negative charge is assumed to start at inﬁnityand end at the negative charge.• Field lines do not cross each other. ( if they did, the ﬁeld at the point ofcrossing will not be uniquely deﬁned.)• A neutral point is a point at which ﬁeld strength is zero. This occurs becauseof cancellation of electric ﬁeld at such a point due to multiple charges.Exercise : Draw ﬁeld lines and show the neutral point for a charge +4Q locatedat (1, 0) and −Q located at (−1, 0).2.3 Electric FluxThe concept of ﬂux is borrowed from ﬂow of water through a surface. The amountof water ﬂowing through a surface depends on the velocity of water, the area of thesurface and the orientation of the surface with respect to the direction of velocityof water.Though an area is generally considered as a scalar, an element of area may beconsidered to be a vector because :• It has magnitude (measured in m2).• If the area is inﬁnitisimally small, it can be considered to be in a plane. Wecan then associate a direction with it. For instance, if the area element liesin the x-y plane, it can be considered to be directed along the z–direction.(Conventionally, the direction of the area is taken to be along the outwardnormal.)2
- 3. SθIn the ﬁgure above, the length of the vector S is chosen to represent the areain some convenient unit and its direction is taken to be along the outwardnormal to the area.We deﬁne the ﬂux of the electric ﬁeld through an area dS to be given by the scalarproductdφ = E · dSIf θ is the angle between the electric ﬁeld and the area vectordφ =| E || dS | cos θ. For an arbitrary surface S, the ﬂux is obtainted by integrating over all the surfaceelementsφ = dφ =SE · dSIf the electric ﬁeld is uniform, the angle θ is constant and we haveφ = ES cos θ = E(S cos θ)Thus the ﬂux is equal to the product of magnitude of the electric ﬁeld and theprojection of area perpendicular to the ﬁeld.3
- 4. SEθUnit of ﬂux is N-m2/C. Flux is positive if the ﬁeld lines come out of the surfaceand is negative if they go into it.Solid Angle :The concept of solid angle is a natural extension of a plane angle to three dimen-sions. Consider an area element dS at a distance r from a point P. Let ˆn be theunit vector along the outward normal to dS.The element of the solid angle sub-tended by the area element at P is de-ﬁned asdΩ =dS⊥r2where dS⊥ is the projection of dSalong a direction perpendicular to r.If α is the angle between ˆr and ˆn,then,dΩ =dS cos αr2rn^^P d ΩαdSSolid angle is dimensionless. However, for practical reasons it is measured interms of a unit called steradian (much like the way a planar angle is measured interms of degrees).The maximum possible value of solid angle is 4π, which is the angle subtendedby an area which encloses the point P completely.Example 1:A right circular cone has a semi-vertical angle α. Calculate the solid angle at theapex P of the cone.Solution :The cap on the cone is a part of a sphere of radius R, the slant length of the cone.Using spherical polar coordinates, an area element on the cap is R2sin θdθdφ,where θ is the polar angle and φ is the azimuthal angle. Here, φ goes from 0 to 2π4
- 5. while θ goes from 0 to α.Thus the area of the cap isdA = 2πR2α0sin θdθ= 2πR2(1 − cos α)Thus the solid angle at P isdΩ =dAR2= 2π(1 − cos α)RP θExercise :Calculate the solid angle subtended by an octant of a sphere at the centre of thesphere. (Ans. π/2)The ﬂux per unit solid angle is known as the intensity.Example 2An wedge in the shape of a rectangular box is kept on a horizontal ﬂoor. Thetwo triangular faces and the rectangular face ABFE are in the vertical plane. Theelectric ﬁeld is horizontal, has a magnitude 8 × 104N/C and enters the wedgethrough the face ABFE, as shown. Calculate the ﬂux through each of the facesand through the entire surface of the wedge.0000000000000000000000000000000000000000000000000000000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111111111111111111111111110.4 m0.3m0.2mEABCDEFSolution :The outward normals to the triangular faces AED, BFC, as well as the normal tothe base are perpendicular to E. Hence the ﬂux through each of these faces is zero.The vertical rectangular face ABFE has an area 0.06 m2. The outward normal tothis face is perpendicular to the electric ﬁeld. The ﬂux is entering through this5
- 6. face and is negative. Thus ﬂux through ABFE isφ1 = −0.06 × 8 × 104= −.48 × 104N − m2/CTo ﬁnd the ﬂux through the slanted face, we need the angle that the normal to thisface makes with the horizontal electric ﬁeld. Since the electric ﬁeld is perpen-dicular to the side ABFE, this angle is equal to the angle between AE and AD,which is cos−1(.3/.5). The area of the slanted face ABCD is 0.1 m2. Thus theﬂux through ABCD isφ2 = 0.1 × 8 × 104× (.3/.5) = +0.48 × 104N − m2/CThe ﬂux through the entire surface of the wedge is φ1 + φ2 = 0.Example 3:Calculate the ﬂux through the base of the cone of radius R.ESolution :The ﬂux entering is perpendicular to the base. Since the outward normal to the cir-cular base is in the opposite sense, the ﬂux is negative and is equal to the productof the magnitude of the ﬁeld and the area of the base, The ﬂux, therefore is, πR2E.Example 4 :Calculate the ﬂux coming out through the curved surface of the cone in the aboveexample.Solution :6
- 7. dlhH rθRConsider a circular strip of radius r at a depth h from the apex of the cone. Theangle between the electric ﬁeld through the strip and the vector dS is π −θ, whereθ is the semi-angle of the cone. If dl is the length element along the slope, the areaof the strip is 2πrdl. Thus,E · dS = 2πrdl | E | sin θWe have, l = h/ cos θ, so that dl = dh/ cos θ. Further, r = h tan θ Substituting,we getE · dS = 2πh tan2θ | E | dhIntegrating from h = 0 to h = H, the height of the cone, the outward ﬂux is| E | H2tan2θ = πR2| E |.Example 5 :A charge Q is located at the center of a sphere of radius R. Calculate the ﬂuxgoing out through the surface of the sphere.dSBy Coulomb’s law, the ﬁeld due to the charge Q is radial and is given on thesurface of the sphere by,E =14πǫ0QR2ˆr7
- 8. The direction of the area vector dS, is also radial at each point of the surfacedS = dSˆr. The ﬂuxφ = E · dS=14πǫ0QR2dSThe integral over dS is equal to the surface area of the sphere, which is, 4πR2.Thus the ﬂux out of the surface of the sphere isφ =14πǫ0QR2· 4πR2=Qǫ0Note that the ﬂux is independent of the radius of the sphere - a cancellationdue to the fact that the surface area of the sphere is proportional to r2whilethe ﬁeld is proportional to 1/r2. Curiously, the result is valid for any arbitrar-ily shaped surface. Consider a cone of solid angle dΩ centered at the charge.This will intersect the arbitrarilyshaped surface in an ellipse whosenormal ˆn makes an angle θ with theoutgoing radial direction ˆr. If thearea of the ellipse is dS,dΩ =dS cos θr2so that the ﬂux through the cone isdΦ =14πǫ0qr2ˆr · dS=14πǫ0qr2ˆr ·r2dΩcos θˆn=14πǫ0qdΩˆn · rcos θ=qdΩ4πǫ0(1)qdSrn^^rdΩThus the total ﬂux throgh the surface isΦ = dΦ =q4πǫ0dΩ =qǫ0(2)8
- 9. One can generalize this to multiple charges since superposition principle holds forthe electric ﬁeld. Suppose the total ﬁeld consists of ﬁelds E1 due to charge q1, E2due to q2 and so on. We haveE · dS =iEi · dS =iqiǫ0One should note that the abovederivation is valid only if the chargesare contained within the volume foronly then the total solid angle be-comes 4π. For charges which areoutside the volume, the ﬂux that en-ters the volume also leaves it andthough the areas which the cone in-tersects are different, the solid anglesare the same. This leads to cancella-tion of the ﬂux and the contributionto the ﬂux from a charge which re-sides outside the volume is zero.rdΩqr^n^n^ dSdS1122GAUSS’S LAW - Integral formThe ﬂux calculation done in Example 4 above is a general result for ﬂux out ofany closed surface, known as Gauss’s law.Total outward electric ﬂux φ through a closed surface S is equal to 1/ǫ0 timesthe charge enclosed by the volume deﬁned by the surface S9
- 10. 01E01E01E0000111101dSMathematicaly, the surface integral of the electric ﬁeld over any closed surface isequal to the net charge enclosed divided by ǫ0E · dS =Qenclosedǫ0(3)• The law is valid for arbitry shaped surface, real or imaginary.• Its physical content is the same as that of Coulomb’s law.• In practice, it allows evaluation of electric ﬁeld in many practical situa-tions by forming imagined surfaces which exploit symmetry of the problem.Such surfaces are called Gaussian surfaces.GAUSS’S LAW - Differential formThe integral form of Gauss’s law can be converted to a differential form byusing the divergence theorem. If V is the volume enclosed by the surface S,SE · dS =V∇ · Edv (A)If ρ is the volume charge density,Q =Vρdv (B)10
- 11. Thus we have∇ · E =ρǫ0(4)Direct Calculation of divergence from Coulomb’s Law :We will use the ﬁeld expressionE(r) =14πǫ0(r − r′)| r − r′ |3ρ(r′)d3rto directly evaluate the divergence of the electric ﬁeld. Since the differentiationis with respect to r while the integration is with respect to r′, we can take thedivergence inside the integral,∇ · E(r) =14πǫ0∇r ·(r − r′)| r − r′ |3ρ(r′)d3rA simple minded calculation of divergence is as follows :∇r ·(r − r′)| r − r′ |3= (∇r · r)1| r − r′ |3+ (r − r′) · ∇r1| r − r′ |3Divergence of r is equal to 3∇r · r =∂∂xx +∂∂yy +∂∂zz = 3∇r1| r − r′ |3= −3r − r′| r − r′ |5Thus it would seem that∇r ·(r − r′)| r − r′ |3= 0i.e. ∇ · E = 0, which violates Gauss’s Law, that ∇ · E = ρ/ǫ0. The problemarises because the function 1/ | r − r′| has a singularity at r = r′. This point hasto be taken care of with care. Except at this point the divergence of the integral isindeed zero.11
- 12. Thus we can shrink the range of in-tegral till it becomes a small spherearound the point r(x, y, z). Sinceρ(x′, y′, z′) is continuous, we mayreplace the density on the surface ofthe small sphere by the value of den-sity at the centre, so that the densityterm can come out of the integral,leaving,∇·E(r) =14πǫ0ρ(r) ∇r·(r − r′)| r − r′ |3d3rr − r’(x,y,z)(x’,y’,z’)Since the divergence with respect to r is being taken of a function which onlydepends on the difference r − r′, we may replace ∇r → −∇′r, which gives∇ · E(r) = −14πǫ0ρ(r) ∇r′ ·(r − r′)| r − r′ |3d3rThe volume integral on the right may be converted to a surface integral using thedivergence theore,∇ · E(r) = −14πǫ0ρ(r)(r − r′)| r − r′ |3· dSwhere the integral is over the surface of the sphere. If the radius of the sphere betaken as r0, r − r′= −r0ˆn and dS = dSˆn, Hence∇ · E = +14πǫ0ρ(r)dSr20=14πǫ0ρ(r) · 4πr20 =ρ(r)ǫ0Example 1: A sphere of radius R contains a continuous charge distribution ρ(r).The electric ﬁeld at a distance r from the centre of the sphere is E = kr3ˆr.(a) Find the charge density.Soln. :∇E =1r2∂∂r(r2Er) =1r2∂∂r(kr5) = 5kr2= ρ/ǫoThus ρ = 5kr2ǫo. (b) Find the total charge contained in the sphere.Soln. :The total charge is obtained by integrating the charge densityR0ρdτ = 5ǫok r24πr2dr = 4πkǫ0R512
- 13. The same result is also obtained by the surface integral of the electric ﬁeld :E · dS = kR3ˆr · (R2sin θdθdφˆr) = 4πkR5=Qǫ0The two results are consistent.Example 2 :Calculate the ﬂux through the shadedarea (face of a cube of side a) when acharge q is located at one of the dis-tant corners from the side.000000000000000000000000000000000000000000000000000000000000000000000000111111111111111111111111111111111111111111111111111111111111111111111111qSolution :If the charge were located at the centre of the cube instead of the corner, the ﬂuxwould have been q/6ǫ0, by symmetry. To use this symmetry consider the givencube to be a part of a bigger cube of side 2a × 2a, as shown, so that the charge qis in the centre of the bigger cube.The ﬂux through each face of thebegger cube is now q/6ǫ0. Be-cause the side of the bigger cubeconsists of four identical faces, theﬂux through one fourth of the face isclearly q/24ǫ0.0000000011111111Applications of Gauss’s LawField due to a uniformly charged sphere of radius R with a charge QBy symmetry, the ﬁeld is radial. Gaussian surface is a concentric sphere of radiusr. The outward normals to the Gaussian surface is parallel to the ﬁeld E at everypoint. Hence E · dS = 4πr2| E |13
- 14. QrRERrQEr >R r < RFor r > R,4πr2| E |=Qr2so thatE =Q4πr2ˆrThe ﬁeld outside the sphere is what it would be if all the charge is concentrated atthe origin of the sphere.14
- 15. For r < R, a fraction r3/R3of thetotal charge is enclosed within thegaussian surface, so that4πr2| E |=1ǫ0Qr3R3The ﬁeld inside isE =Q4πǫ0rR3ˆrEinarbitraryunitsr--->Field due to sphere with charge density ρRExercise :Find the electric ﬁeld both inside and outside a spherical shell of radius R carryinga uniform charge Q.Example 3 :Find the electric ﬁeld inside a sphere of radius R which carries a charge densityρ = kr where r is the distance from the origin and k is a constant.Solution :By symmetry the ﬁeld is radial. Take the gaussian surface to be a sphere of radiusR. The ﬂus is 4πr2| E |.15
- 16. rRThe charge enclosed by the gaussian surface isQ =r0ρ(r)d3r =r0ρ(r)4πr2dr= 4πkr0r3dr= πkr4ThusE =1ǫ0kr24ˆr(what is the dimension of k ?)Exercise : A hollow spherical shell carries a charge density ρ = k/r2fora ≤ r ≤ b. Calculate the ﬁeld at all points. (Ans. For r < a, ﬁeld is zero, fora < r < b, | E |= k(r − a)/ǫ0r2, and for r > b, | E |= k(b − a)ǫ0r2.)Field due to an inﬁnite line charge of linear charge density λGaussian surface is a cylinder of radius r and length L.By symmetry, the ﬁeld has the same magnitude at every point on the curved sur-face and is directed outwards. At the end caps, E is perpendicular to dS every-where and the ﬂux is zero. For the curved surface, E and dS are parallel,16
- 17. dS1dS2EL++++++++++++++++E · dS = | E | .2πrL=Qǫ0=λLǫ0ThusE =λ2πǫ0rˆρwhere ˆρ is a unit vector perpendicular to the line,directed outward for positive linecharge and inward for negative line charge.Exercise :Find the electric ﬁeld both inside and outside a long cylinder of radius R carryinga uniform volume charge density ρ.(Hint : Take the gaussian surface to be a ﬁnite concentric cylinder of radius r (withr < R and r > R), as shown)+ + + + + + +++ +3 Field due to an inﬁnite charged sheet with surface charge density σChoose a cylindrical Gaussian pillbox of height h (with h/2 above the sheet andh/2 below the sheet) and radius r.17
- 18. dSE+++++++ + ++++++ +++++ +++++++LσdSEErdSThe amount of charge enclosed is area times the surface charge density, i.e., Q =πr2σ. By symmetry, the ﬁeld is directed perpendicular to the sheet, upward atpoints above the sheet and downward for points below. There is no contributionto the ﬂux from the curved surface. The ﬂux from the two end faces is πr2| E |each, i.e. a total outward ﬂux of 2πr2| E |. Hence2πr2| E |=Qǫ0=πr2σǫ0so thatE =σ2ǫ0ˆnwhere ˆn is a unit vector perpendicular to the sheet, directed upward for pointsabove and downwards for points below (opposite, if the charge density is nega-tive).Exercise :Find the electric ﬁeld in the region between two inﬁnite parallel planes carryingcharge densities +σ and −σ.18
- 19. +++++++++σ −σExercise :A very long cylinder carries a charge density ρ = kr, where r is the distance fromthe axis of the cylinder. Find the electric ﬁeld at a distance r < R. (Ans.(1/3ǫ0)kr2ˆr)Example 4 :Two spheres of radius R each overlap such that the distance between their centresis separated by a distance 2R − s. Show that the ﬁeld in the overlapping region isconstant.Solution :O O’sθφPQThe ﬁgure shows the ﬁeld at a point P in the overlap region due to the two spheres.Taking the expression for ﬁeld at a point inside the sphere and resolving into x-and y- components, The x- components reinforce while the y-components are in19
- 20. the opposite directions. It can be seen,Ey =Q4πǫ01R3(OP sin θ − O′P sin φ)Using the property of triangles (the ratio of the sides is equal to the ratio of thesine of opposite angles a/ sin A = b/ sin B) the y-component is seen to vanish.The x-component of the ﬁeld isEx =Q4πǫ01R3(OP cos θ + O′P cos φ)=Q4πǫ01R3(OQ + O′Q)=Q4πǫ01R3sThus the ﬁeld depends only on the distance between the centres and is constant.Example 5 :A sphere of radius R has a cavity of radius a inside it. The sphere has uniformcharge density ρ spread over its volume. Show that the ﬁeld inside the cavity isconstant.Solution :One can use superposition princi-ple to solve this problem. Fill upthe cavity with equal and oppositecharge distribution. The problenmthen is equivalent to the ﬁeld due toa sphere of radius R and charge den-sity ρ and a smaller sphere of radiusa, but with a charge densiuty −ρ.RadrOPO’We calculate the ﬁelds due to these two spheres at r (with respect to O, the centre20
- 21. of the larger sphere, The ﬁeld due to larger sphereE1 =Q4πǫ0rR3=ρ4πǫ04π3R2 rR3=ρ3ǫ0rBy identical argument, the ﬁeld due to smaller sphere (the point P is at r − d withrespect to the centre of the smaller sphere),E2 = −ρ3ǫ0(r − d)Adding,E =ρ3ǫ0d = constant21

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