Electrc Potental/Energy Phys0 General Physcs II Electrc Potental Topcs Electrc potental energy and electrc potental Equpotental Surace Calculaton o potental rom eld Potental rom a pont charge Potental due to a group o pont charges, electrc dpole Potental due to contnuous charged dstrbutons Calculatng the eld rom the potental Electrc potental energy rom a system o pont charge Potental o a charged solated conductor Electrc potental energy The electrc orce s mathematcally the same as gravty so t too must be a conservatve orce We wll nd t useul to dene a potental energy as s the case or gravty Recall that the change n the potental energy n movng rom one pont to pont s the negatve o the work done by the electrc orce U U = U U = Work done by the electrc orce U = F ds Snce F=qE, U = q E ds Electrc potental energy U U = U U = Work done by the electrc orce W = F d F = qe W = qed = qedcosθ
Electrc potental Equpotental Suraces Electrc Potental = Potental energy / unt charge V= = U/q Electrc Potental derence = Potental energy change/ unt charge V V = V V = (U( U )/q= U/q V V = V V = W/q The potental derence s the negatve o the work done per unt charge by an electrc eld on a postve unt charge when t moves rom pont to pont The SI unts o V are Joules/ Coulomb or Volt E s N/C or V/m Frst draw the eld lnes, then nd surace perpendcular to these lnes Equpotental Suraces Equpotental Suraces Unorm E eld E = E x, E y = 0, E z = 0 V = E x d V = constant n y and z drectons Pont charge (concentrc shells) Two pont charges (ellpsodal concentrc shells)
U = U b U a = Work done by the electrc orce = F ds V= = U/q Electrc orce s a conservatve orce Thereore ndependent o path V = V V = Eds V = V V = Eds We are ree to choose V to be 0 at any locaton Normally V s chosen to be 0 at nnty or a pont charge V = Eds Examples What s the electrc potental derence or a postve charge movng n an unorm electrc eld? a b V = a d V = Ed b E E b a x drecton E ds = E dx = E( xb xa) d U = q V U = qed Examples In a 9 volt battery, typcally used n IC crcuts, the postve termnal has a potental 9 v hgher than the negatve termnal I one mcrocoulomb o postve charge lows through an external crcut rom the postve to negatve termnal, how much has ts potental energy been changed? q U V = = Vb Va = 0 9 q U = 9q = 9V x 0 6 C U = 9V x 0 6 Joules U = 9 mcrojoules Potental energy s lower by 9 mcrojoules 3
Examples A proton s placed n an electrc eld o E=0 5 V/m and released Ater gong 0 cm, what s ts speed? Use conservaton o energy a b V = V b V a = Ed U = q V U K = 0 K = U (/)mv = q V = qed E = 0 5 V/m d = 0 cm K = (/)mv v = v = qed m 6 0 v = 4 0 C 0 9 5 V m 3 67 0 6 m s kg m Electrc potental due to a pont charge V V = E ds V V = R 0 V = k q R E dr = kq kq E = r We choose V=0 at r = dr = kq r r R kq V = R V s a scalar, not a vector V s postve or postve charges, negatve or negatve charges V = k = 4πε 0 q 4πε 0 r Electrc potental or a postve pont charge V (r) = kq /r r = x y Electrc potental due to a postve pont charge Hydrogen atom What s the electrc potental at a dstance o 059 A rom the proton? r r = 059 A V = kq/r = 899*0 0 N m / /C * 6*0 9 C/059*0 0 m V = 7 J/C = 7 Volts 4
V Electrc potental due to pont charges = q 4πε 0 r Electrc potental due to pont charges For many pont charges, the potental at a pont n space s the sum V= Σ kq /r For many pont charges, the potental at a pont n space s the sum V= Σ kq /r q q V = k r r y r r p r 3 3 x q r 3 3 Electrc potental due to an electrc dpole q V = 4πε 0 r Ways o Fndng V Drect ntegraton Snce V s a scalar, t s easer to evaluate V than E Fnd V on the axs o a rng o total charge Q Use the ormula or a pont charge, but replace q wth elemental charge dq and ntegrate Pont charge V = kq/r For an element o charge dq, dv = kdq/r r s a constant as we ntegrate V = kdq/r = kdq/(z R ) / =k/(z R ) / dq = k/(z R ) / Q Ths s smpler than ndng E because V s not a vector V = kq/r 5
Lne o Charge What s the electrc potental o a unormly charged crcular dsk? More Ways o ndng V and E Use Gauss Law to nd E, then use V= Eds to get V V = Eds Suppose E y = 000 V/m What s V? V = 000 y Calculatng the Feld rom Potental More generally, I we know V, how do we nd E? dv= E ds ds = dx j dyk dz and dv = E x dx E y dy E z dz E x = dv/dx, E y = dv/dy, E z = dv/dz So the x component o E s the dervatve o V wth respect to x, etc I E x = 0, then V = constant n that drecton Then lnes or suraces on whch V remans constant are called equpotental lnes or suraces (Unorm eld) V = Ed 6
Electrc Potental Energy o a system o pont charges Potental o a charged solated conductor Start out wth a unorm electrc eld wth no excess charge on conductor Electrons on surace o conductor adjust so that: E=0 nsde conductor Electrc eld lnes are perpendcular to the surace Suppose they weren t? 3 Is the surace an equpotental? V = V V = Eds How does a conductor sheld the nteror rom an exteror electrc eld? Start out wth a unorm electrc eld wth no excess charge on conductor Electrons on surace o conductor adjust so that: Delectrc Breakdown: Applcaton o Gauss s Law I the electrc eld n a gas exceeds a certan value, the gas breaks down and you get a spark or lghtnng bolt the gas s ar In dry ar at STP, you get a spark when E = 3*0 6 V/m To examne ths we model the shape o a conductor wth two derent spheres at each end: E=0 nsde conductor Electrc eld lnes are perpendcular to the surace Suppose they weren t? 3 Does E = σ/ε 0 just outsde the conductor 4 Is σ unorm over the surace? 5 Is the surace an equpotental? V = constant on surace o conductor Radus R 6 I the surace had an excess charge, how would your answers change? 7
The surace s at the same potental everywhere, but charge densty and electrc elds are derent For a sphere, V= q/(4π ε 0 r) and q = 4πr4 σ, then V = (σ/( ε 0 )*r Snce E = σ/ ε 0 near the surace o the conductor, we get V=E*r Snce V s a constant, E must vary as /r and σ as /r Hence, or suraces where the radus s smaller, the electrc eld and charge wll be larger Ths explans why: V = constant on surace o conductor Radus R Sharp ponts on conductors have the hghest electrc elds and cause corona dscharge or sparks A metal slab s put n a unorm electrc eld o 0 6 N/C wth the eld perpendcular to both suraces Draw the approprate model or the problem Show how the charges are dstrbuted on the conductor Draw the approprate pll boxes What s the charge densty on each ace o the slab? Apply Gauss s Law EdA = q n /ε 0 Pck up the most charge wth charge tester rom the ponty regons o the non sphercal conductor Slab o metal In a unorm Electrc eld Slab o metal In a unorm Electrc eld E = 0 6 N/C Gaussan Pll Box Evaluate EdA = q n /ε 0 Let sde o slab E*A 0*A = Aσ/ε 0, E = σ /ε 0, σ = 0 6 N/C *0 C /Nm = 0 5 N/m Rght sde o slab 0*A E*A = Aσ/ε 0, E = σ /ε 0, σ = 0 6 N/C *0 C /Nm = 0 5 N/m (note that charges arranges themselves so that eld nsde s 0) 8