Electricity and Magnetism--Electric Potential Energy, Gauss's law

Topics: Energy associated with an electric field in!) 8.022 (E&M) – Lecture 3 Electric potential Gauss’s law in differential form … and a lot of vector calculus… (yes, aga2 Last time… What did we learn? Energy of a system of charges 4 S E dA QπΦ= =∫ 􀁇􀁇 i􀁶 2ˆ || qF QE r q r = = 􀁇 􀁇 1 1 1 2 jNiN ij i j ijji qqU r == = = ≠ = ∑∑ G. Sciolla – MIT 8.022 – Lecture 3 Electric field Gauss’s law in integral form: Derived last time, but not rigorously… encl 13 Gauss’s law 1 solid angle dΩfrom dΦS=dΦS1 Æ ΦS =ΦS1=4πQ 4S QπΦ= =∫ 􀁇􀁇 i􀁶 S θˆn R S1r2 1 2 ˆ ˆ( )( )S qd qdr Φ= = Ω = Ω 􀁇􀁇 i 2 2 ˆˆˆ ˆ( ) ( )S qd r n qd qdR θ θ ΩΦ= = = Ω = Ω 􀁇􀁇 ii i Q NB: E~1/r2the r2 would not cancel!!! G. Sciolla – MIT 8.022 – Lecture 3 Consider charge in a generic surface S Surround charge with spherical surface Sconcentric to charge Consider cone of charge to surface S through the little sphere Electric flux through little sphere: Electric flux through surface S: is valid for ANY shape S. encl EdA E dA r r d r cos cos R d r n E dA Gauss’s law only because . If E ~ anything else, 4 Confirmation of Gauss’s law 2 ˆ0 Q rE r ⎧⎪ =⎨ ⎪⎩ 􀁇 Does E~1/r2φ~1/r? (D24)+ + ++ + + + + G. Sciolla – MIT 8.022 – Lecture 3 Electric field of spherical shell of charges: Can we verify this experimentally? o utside the shell in sid e the shell • Charge a spherical surface with Van de Graaf generator • Is it charged? (D7 and D8) • Is Electric Field radial? , eg: Neon tube on only when oriented radially • (D29?) 25 Confirmation of Gauss’s law (2) ll positiE= 0 E> 0 Demo D26 i iiil> 0 Æ i+ + + E>0 E=0 G. Sciolla – MIT 8.022 – Lecture 3 Cylindrical shevely charged Gauus tells us that inside outside Can we verify this experimentally? Charge 2 conductve spheres bynducton outside the cylnder: one sphere wil be + and the other will be -: it works because Eoutside Try to do the same inside inside cylinder nothing happens because E=0 (explainnduction on the board) 6 Energy stored in E: Squeezing charges… x y ius r-dr? ( ) 2 2 2 2 2 2 1; 0 2 1 4 2 2 2 F QP E EA A A Q QE E E r r QPE r r r σ σσ σ π σ = = = = = = → = →= = = = 2 2 2 2 ( ) (2 )(4 ) 2 4 ) dW Fdr r dr dV dV πσ π πσ π = = = = = 2 8 EdW dVπσ π⇒ = G. Sciolla – MIT 8.022 – Lecture 3 Consider a spherical shell of charge of radius r How much work dW to “squeeze” it to a radGuess the pressure necessary to squeeze it: We can now calculate dW: outside inside surface QE πσ (where PA dr r dr created in dr Remembering that E =4 37 Energy stored in the electric field Work done on the system: Where does the energy go? Æ iiÆ is the energy density of the electric field E Energy is stored in the E field: 2 8 E u π = 2 8 EdW dVπ = 2 8 EU π = ∫ G. Sciolla – MIT 8.022 – Lecture 3 We do work on the system (dW): same sign charges have been squeezed on a smaller surface, closer together and they do not like that… We created electric field where there was none (between r and r-dr) The electrc field we created must be storng the energy Energy is conserved dU = dW NB: integrate over entire space not only where charges are! Example: charged sphere Entire space dV 8 Electric potential difference 1 to r2: W12 /Æ define a quantity that is independent of q and just describes the propertiPhysical interpretation: φ12 1 to P2 Units: 2 V r1r2 q Q ds􀁇 2 2 2 12 1 1 1IW F F= • = − • = −∫ ∫ ∫ 􀁊􀁇􀁇 􀁇􀁇 􀁇 􀁇i 212 12 1 W E ds q φ ≡ ∫ 􀁊􀁇 􀁇i Electric potential difference between P1 and P2 G. Sciolla – MIT 8.022 – Lecture 3 Work to move q from rdepends on the test charge q es of the space: is work that I must do to move a unit charge from Pcgs: statvolts = erg/esu; SI: Volt = N/C; 1 statvolts = “3” 10Coulomb ds ds q E ds = −49 Electric potential φ12 1 2: we need 2 points! 1Application 1: Calculate φApplication 21 and P2: Æ ( )r rφ ∞ =− ⇐∫ 􀁇 􀁊􀁇􀁇 􀁇i 2 12 2 11 2 1 () ( )r r q qE ds P P r r φ φ φ=− = − = −∫ 􀁊􀁇 􀁇i 2() r r q qr r r φ ∞ ∞ =− =− =∫ ∫ 􀁇 􀁊􀁇􀁇 􀁇i G. Sciolla – MIT 8.022 – Lecture 3 The electric potential difference is defined as the work to move a unit charge between Pand PCan we define similar concept describing the properties of the space? Yes, just fix one of the points (e.g.: P=infinity): (r) created by a point charge in the origin: : Calculate potential difference between points PPotential difference is really the difference of potentials! Potential E ds E ds dr Potentials of standard charge distributions 􀁇 rThe potential created by a point charge is φ() =q r Æ Given this + superposition we can calculate anything! N􀁇 ir Potential of N point charges: φ() =∑q i=1 ri 􀁇 ρdV r 􀁇 σdA Potential of charges in a volume V: φ() =∫Vr r Potential of charges on a surface S: φ() =∫Sr􀁇 λdlrPotential of charges on a line L: φ() =∫Lr G. Sciolla – MIT 8.022 – Lecture 3 10 5 Some thoughts on potential Why is potential useful? Isn’t E good enough? Potential is a scalar function Æ much easier to integrate than electric field or force that are vector functions When is the potential defined? Unless you set your reference somehow, the potential has no meaningUsually we choose φ(infinity)=0This does not work always: e.g.: potential created by a line of chargesCareful: do not confuse potential φ(x,y,z) with potential energy of asystem of charges (U) 1 iN jNqq== ij Potential energy of a system of charges: U= ∑∑2 i=1 j=1 rijwork done to assemble charge configuration ji≠ N􀁇 irPotential: work to move test charge from infinity to (x,y,z) φ() =∑q i=1 ri G. Sciolla – MIT 8.022 – Lecture 3 11 Energy of electric field revisited where φ(rjj at the location of qj (rj) Taking a continuum limit: φ(infinity)=0 1 1 1 2 jNiN ij i j ijji qqU r == = = ≠ = ∑∑ 1 1 ()2 2 i j j j ji i jiij qU q q r r φ ≠ ≠ = =∑∑ ∑ 21 ()2 8 EU rdV dVρφ π = =∫ ∫ G. Sciolla – MIT 8.022 – Lecture 3 12 Energy stored in a system of charges: This can be rewritten as follows: ) is the potential due to all charges excepted for the qNB: this works only when Volume Entire with space charges 6 φ φConnection between φand E E φ=−∇ 􀁇 ~ ( )rdr r d s rφ +=− −∫ 􀁇 􀁇 􀁇 􀁊􀁇 􀁇􀁇i i ( ), , , ,d dx dy dz dz dr x y z x φ φ φφ φ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ = + + ≡ • •⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ 􀁇 G. Sciolla – MIT 8.022 – Lecture 3 13 Consider potential difference between a point at r and r+dr: The infinitesimal change in potential can be written as: Useful info because it allows us to find E given Good because is much easier to calculate than E EdEr d􀁊􀁇 􀁇 xdydy z φ φ φ ≡∇ 1d problem: Æ 2d problem: Æ Getting familiar with gradients… ˆ() ffx x x ∂∇ ≡ ∂ ˆ ˆ(, ) , f f x yx y ⎛ ⎞∂ ∂ ∂ ∂∇ ≡ + ≡⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ G. Sciolla – MIT 8.022 – Lecture 3 14 The derivative df/dx describes the function’s slope The gradient describes the change of the function and the direction of the change The interpretation is the same, but in both directions The gradient points in the direction where the slope is deepest f f fxy x y 7G. Sciolla – MIT 8.022 – Lecture 3 15 Given the potential φ Visualization of gradients ˆ ˆ( , ) ) sin( )xy x x yyφ∇ = + grad φφ (x,y) Æ E=-gradφ points downhill (x,y)=sin(x)sin(y), calculate its gradient. cos( sin cos yx The gradient always points uphill Same potential φ (x,y)=sin(x)sin(y) φ (x,y) Æ G. Sciolla – MIT 8.022 – Lecture 3 16 Visualization of gradients: equipotential surfaces NB: since equipotential lines are perpendicular to the gradient equipotential lines are always perpendicular to E 8S20 S2 S10 S1 Divergence in E&M (1) 1 and S2 with Snew the littl11 2 2 1 1 2 2 1 1 22 S ew new S ew S S new S S E− − Φ= = + = − + − = − = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 􀁇 􀁇 􀁇􀁇 􀁇 􀁇 i i i 􀁇 􀁇 􀁇 􀁇􀁇 􀁇 􀁇 􀁇 i i i i 􀁇 􀁇􀁇 􀁇 i i 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 􀁶 S Snew S2 S1 S1new S2new dA dA G. Sciolla – MIT 8.022 – Lecture 3 17 Consider flux of E through surface S: Cut S into 2 surfaces: Se surface in between S SnS S S nEdA EdA EdA dA EdA EdA EdA EdA EdA Φ+Φ Divergence Theorem Æ Æ 1 1 1 i i i iSi i i iSii i i i EdA V V = = = = = = Φ= = ∫∑ ∑ ∑∫ 􀁇􀁇 i􀁇􀁇 i 􀁶 􀁶 0 S V EdA E V→ ∇ ≡ ∫ 􀁇􀁇 i􀁇 i 􀁶 S V E=∇∫ ∫ 􀁇􀁇 􀁇 i i Divergence Theorem 1 ( )i Vi V E = Φ= ∇ → ∇∑ ∫ 􀁇 􀁇 i i G. Sciolla – MIT 8.022 – Lecture 3 18 Let’s continue splitting into smaller volumes If we define the divergence of E as largeN largeN largeN EdA Φ = lim EdA dV (Gauss’s Theorem) largeN EdV 9Gauss’s law in differential form First Maxwell’s equations ρ ( 4 )4 4 S V V S V EdA EdV E dV EdA Q dV πρ π ⎧ =∇⎪ → ∇ −⎨ = =⎪⎩ ∫ ∫ ∫∫ ∫ 􀁇􀁇 􀁇 i i 􀁇 i􀁇􀁇 i 􀁶 􀁶 4E πρ∇ = 􀁇 i G. Sciolla – MIT 8.022 – Lecture 3 19 Simple application of the divergence theorem: This is valid for any surface V: Comments: Given E, allows to easily extract charge distribution = 0 π ρ What’s a divergence? Since ∆zÆ0 Φx and Φy P ~ [ (, , ) (, , )2 2z z z toottom z z + ∆ ∆+ − −∫ 􀁇􀁇 i 0 1( ) [ (, , ) (, , )]2 2 z z z z z Fz z z z∆→ ∂∆ ∆+ − − =∆ ∂yx x y FF x y ∂∂ ∆∂ ∂ x y z ∆y ∆x ∆z G. Sciolla – MIT 8.022 – Lecture 3 20 Consider infinitesimal cube centered at P=(x,y,z) Flux of F through the cube in z direction: Similarly for pbFdA xyFxyz Fxyz ∆Φ = ∆∆ lim xyz Fxyz Fxyz xyz ∆Φ = ∆∆∆ ∆∆∆ and xy z xyz ∆Φ =∆∆∆ Φ =∆∆∆ 10Divergence in cartesian coordinates mean? 0 S V FdA F V→ ∇ ≡ ∫ 􀁇􀁇 i􀁇 i 􀁶 0 0 0 0 0 0 ( ) S x y z yx z x y z yx z FdA F V FF F x y z FF F x y z ∆→ ∆→ ∆→ ∆→ ∆→ ∆→ ∇ ≡ ∂∂ ∂+ +∂ ∂ ∂ = ∂∂ ∂ = + + ∂ ∂ ∂ ∫ 􀁇􀁇 i􀁇 i 􀁶 divergence: easy to calculate! G. Sciolla – MIT 8.022 – Lecture 3 21 We defined divergence as But what does this really lim lim lim xy z xy z ∆∆∆ ∆∆∆ This is the usable expression for the Application of Gauss’s law in differential form Problem3 2 4 4ˆ ˆ( ) for rR 3 3 KEr Krr Er r ππ= 􀁇 􀁇 Hint: what connects E and ρ4 ) 4 () enclS EdA Q E π πρ = ∇ = ∫ 􀁇􀁇 i 􀁇 i 􀁶 4yx z E KE EE x y z π∂ ⎧∂ ∂∇• ≡ + + = =⎨∂ ∂ ∂ ⎩ 􀁇􀁇 Æ R with constant charge density K G. Sciolla – MIT 8.022 – Lecture 3 22 : given the electric field E(r), calculate the charge distribution that created it Rr ? Gauss’s law. (integral formdifferential formIn cartesian coordinates: when rR Sphere of radius 11Next time… G. Sciolla – MIT 8.022 – Lecture 3 23 Laplace and Poisson equations Curl and its use in Electrostatics Into to conductors (?) 12