Problem Set #3: 2.11, 2.15, 2.21, 2.26, 2.40, 2.42, 2.43, 2.46 (Due Thursday Feb. 27th)

Transcription

1 Chapter Electrostatics Problem Set #3:.,.5,.,.6,.40,.4,.43,.46 (Due Thursday Feb. 7th). Coulomb s Law Coulomb showed experimentally that for two point charges the force is - proportional to each of the charges, - acts in the direction along the line joining charges and - falls of as inverse square of the distance, i.e. In a fixed coordinate system (and in SI units), F k q q ˆr. (.) r F q q r r (r r ) r r q q r r 3 (r r ). (.) It convenient to define an electric field of the second point charge as such that E q r r 3 (r r ) (.3) F q E. (.4) Since E is a vector field it must have a certain direction and a magnitude in every point in space and time. One can always think of electric field as a functions of spatial r (in the first part of the course) and also temporal t (in the second part of the course) coordinates even if the dependence is often suppressed for brevity of notations. 6

2 CHAPTER. ELECTROSTATICS 7.. Continuum distribution If the system consists of N charges q,q...q N at positions r, r...r N,then from the additivity of forces we get additivity of electric fields E(r) N q i r r i 3 (r r i) (.5) or in continuum limit E(r) ρ(r ) r r r r 3 d3 r, (.6) where ρ(r) is charge density. For example, ρ(r) q 0 δ (3) (r r 0 ) for a point charge N ρ(r) q i δ (3) (r r i ) for a collection of point charges ρ(r) λδ(x)δ(y) for a line of charges along z axis ρ(r) σδ(z) for a plane of charges at z 0 Then the electric field for these configurations of charges can be determined by applying equation (.6): Point charge E(r) q 0 δ (3) (r r 0 ) r r r r 3 d3 r q 0 r r 0 r r 0 3 (.7) Collection of charges E(r) N N q i q i δ (3) (r r i ) r r r r 3 d3 r r r i (.8) r r i 3

3 CHAPTER. ELECTROSTATICS 8 Line of charges (along z axis) E(r) λδ(x )δ(y ) r r r r 3 d3 r λ xˆx + yŷ +(z z )ẑ (x + y +(z z ) dz ) 3/ λ xˆx + yŷ + z ẑ (x + y + z dz ) 3/ λ sŝ (s + z dz ) 3/ [ ] λ z + ŝ s s + z [ŝ λ s + ŝ ] λ s π sŝ (.9) Plane of charges (at z 0) E(r)... σ. (.0) We will prove this by a different method shortly... Gauss s law Take a point charge q and enclose arbitrary smooth surface around it. The electric field at the surface is E(r) q rˆr (.) where the origin is chosen to be at the location of the charge. If ˆn is a unit vector normal to the enclosing surface, then E ˆnda q rˆr ˆnda q cos(α)da. (.) r For the spherical coordinates where the differential solid angle r dωcos(α) da (.3) dω sin θdθdφ

4 CHAPTER. ELECTROSTATICS 9 and we obtain the Gauss s law for a point charge in integral form E ˆnda q ˆr ˆnda q dω q. (.4) r For a distribution of charges the Gauss s law (in the integral form) is given by E ˆnda ρ(r)d 3 r. (.5) From the divergence theorem, i.e. A ˆnda Ad 3 r (.6) we get Ed 3 x ρ(r)d 3 r, (.7) which must be true for an arbitrary enclosing surface. Therefore the integrands must be equal, i.e. E ρ(r). (.8) This the Gauss s law in differential form which is one of four Maxwell equations. For example, we can apply the Gauss law to a surface charge density (xy surface at z 0),then E ẑ dx dy E ẑ dx dy σδ(z)dz (.9) top bottom or [(E E ) ẑ σϵ0 ] in agreement with (.0).. Scalar Potential. (.0) z0 It looks as if the electric field E(r) is a vector quantity, however in electrostatics (first part of the course) it is actually a gradient of a scalar quantity. This was shown by an experimental observation that the electric field has a vanishing curl, i.e. E 0. (.)

5 CHAPTER. ELECTROSTATICS 30 But because ( V ) 0 (.) it is tempting to define a scalar potential V such that E V. (.3) If we apply the new definition of the electric potential to Coulomb s law, then V (r) ρ(r ) r r r r 3 d3 r ρ(r ) r r d3 r (.4) or V (r) where we also used the identity ρ(r ) r r d3 r + C (.5) r r r r 3 r r. (.6) (Derive it for yourself!) Moreover from the Gauss s Law (in differential form) we get the Poisson Equation V ρ (.7) or in the absence of charges at a Laplace Equation V 0. (.8) Since ( r d3 r ) ( ) ˆn da r dω 4π r r r we can formally write, 4πδ(r)d 3 r (.9) r r 4πδ(r r ) (.30) which is the Poisson equation of a unit point charge at r. As an example let us calculate the electric fields due to a spherical shell of radius R centered in the origin with uniform surface charge density σ, i.e. ρ(r) δ(r R)σ (.3)

6 CHAPTER. ELECTROSTATICS 3 We could do it directly using Gauss s law E ˆnda Ed 3 r to show that the electric field E { 0 for r<r σ R ˆr for r>r r ρ(r)d 3 r (.3) (.33) or we could first find the electric potential using (.5) V (r) ρ(r ) r r d3 r + C (.34) and then use the definition (.3) to find the electric field or V(r) σ σr σr σr σr z σr z π π 0 π 0 0 R + z Rz cos θ R sin θ dθ dφ R + z Rz cos θ sin θ dθ R + z Rzt dt [ ] R + z Rzt Rz ( R + z +Rz + ) R + z Rz ( (R z) + V (r) { σr (R + z) ). (.35) ˆr for r R σr ˆr for r R. (.36) r Then the electric field is given by (.3) and thus(.36) agrees with (.33)..3 Work and Energy The electric potential measured in [ ] kg m [V ] C s (.37)

7 CHAPTER. ELECTROSTATICS 3 and the potential energy measured in [ ] kg m [J] [V C]. (.38) Evidently the exact values of V cannot be observed, but the differences in V are related to E which is a physical quantity. With this respect the electric potential is similar to potential energy. For example, the work done to move charge from point A to point B in electric field is B B B W F dl q E dl q V dl q(v B V A ) (.39) A A A regardless of the path of integration. It should also be clear that the total work over a closed path is always zero. Then one can apply the Stoke s theorem for an arbitrary area, E dl ( E) ˆnda (.40) to confirm that E 0. Thus, one can think of qv as a potential energy of a charge in electric field. What about a distribution of many charges? If we are to bring an additional charge to a collection of n charges,thentheworkthatneeds to be done is q n W q n V n s q i r i r n. (.4) If we are to assemble from scratch a charge distribution made out of multiple charges the total potential energy would be W j<i q i q j r i r j j q i q j r i r j (.4) or in continuum limit W 8π ρ(r)ρ(r ) r r d3 r d 3 r V (r) V (r)d 3 r. (.43) After integration by parts (and neglecting the boundary term) we get W ϵ 0 V d 3 r ϵ 0 E d 3 r. (.44)

8 CHAPTER. ELECTROSTATICS 33 Then we can identify the integrant with energy density w E (.45) Note that the energy density is always positive (unlike potential energy (.4)) which is due to (infinite and positive) self-energy contribution not included in (.4)..4 Conductors Perfect conductor contains an unlimited supply of point charges (microscopically speaking the electrons) that are free to move. As was already mentioned the charges are moving according to Lorentz force law which reduces in electrostatics to F q (E + v B). (.46) F qe. (.47) This means that if there would be a non-zero net electric field in a conductor the free charges would move (which is exactly what happens in electric circuits). In the context of electrostatics it also implies that the electric field inside conductors should be zero E conductor 0 (.48) and is orthogonal to the surface at the surface. Thus if you bring acharge near a perfect conductor it will induce charges (induced charges) on the surface of the conductor to make the electric field vanish inside a conductor and perpendicular at the surface of the conductor. Moreover, according to Gauss s law the charge density inside conductors is zero ρ conductor 0 (.49) nevertheless the charge can reside on the surface of a conductor. It also follows (directly form the definition) that the electric potential cannot vary across the same conductor V conductor const. (.50)

9 CHAPTER. ELECTROSTATICS 34 Consider an uncharged spherical conductor centered at the origin with a cavity of arbitrary shape. What is the electric field outside of the sphere if there is a charge q inside the cavity? It turns out the the solution is given by E q ˆr. (.5) r This can be shown by solving for electric potential inside of the sphere using the fact that the electric potential must vanish everywhere in the conductor. This can only happen if the charge +q induces a charge q on the surface of the cavity distributed such that the filed in the conductor vanishes. But since the total charge in the conductor is zero, there must be a charge +q distributed uniformly on the surface of the sphere. This implies that the the shape of the cavity as well as the precise location of the charge is hidden from the outside observer. More generally one can consider a collection of perfect conductors. The potential of i th conductor can be written as V i p ij q j (.5) j where p ij depends only on the geometry of the conductors. We can invert the equations to obtain q i C ij V j (.53) j where C ii are called capacitances and C ij (for i j) are called coefficients of induction. Then the potential energy of the system of conductors is or W W q i V i q i V i C ij V i V j (.54) j p ij q i q j (.55) Note that variations of W with respect to V i s or q i s can be used to obtain C ij or p ij. For a surface charge density σ on the surface of a conductor one can integrate the Gauss s law to obtain the electric field outside of the conductor j E σ ˆn (.56) ϵ

10 CHAPTER. ELECTROSTATICS 35 or in terms of electric potential σ V n. (.57) This means that the electric field is discontinues jumping form E inside 0just inside the conductor to E outside σ ˆn just outside the conductor. Then when calculating the force on the surface charge distribution one should average the two, i.e. F σe average σ E inside E outside σ ˆn. (.58) Note that the force is always pointing outward as one might have expected.