Kinetic energy, work, and potential energy. Work, the transfer of energy: force acting through distance: or or

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ENERGETICS So fa we have been studying electic foces and fields acting on chages. This is the dynamics of electicity. But now we will tun to the enegetics of electicity, gaining new insights and new methods fo analyzing poblems just as we did in classical mechanics in the fist semeste.

Electic Potential This chapte is about Electic Potential. We will appoach this topic as follows: Fom classical mechanics, a quick eview: 1. Kinetic enegy, wok, and potential enegy. 2. Mass-sping system, and consevation of total enegy. 3. Gavitational potential enegy, U G, nea Eath (1D). Then, on to electicity: 1. Electic potential enegy, U E, in unifom E (1D). 2. Intoduce electic potential, V E. 3. Find U E and U G in 3D. 4. Solve poblems using electic potential.

Kinetic enegy, wok, and potential enegy Kinetic enegy, the enegy of motion: K = ½ mv 2, fo any given mass. Wok, the tansfe of enegy: foce acting though distance: W = Fd W = F d = Fd cos(θ) o o F d θ W = F ds = F cos(θ) ds Potential enegy, U. Enegy of position: stoed in a system as a esult of wok. Also known as pe-integated wok. Defined only fo consevative systems (those without dissipative foces such as fiction).

Calculating potential enegy The wok-enegy theoem states that wok done by a system as it moves an object fom a to b changes its potential enegy: W = U U = ΔU Example, sping obeying Hooke s Law: F = -kx ab a b Wa b = U a U b = ( U b U a ) = Δ U W x 2 = Fdx = kxdx x 1 x x 2 1 = 1 2 kx 2 2 + 1 2 kx 2 1 = U 2 + U 1 U = 1 kx 2 2 Enegy consevation: E total = K + U = ½ mv 2 + ½ kx 2 = const

Gavitational potential enegy nea Eath s suface The gavitational foce field is nealy unifom and pointed downwad, with a constant foce: F = -mg Wok-enegy theoem U = mgy W y = Fdy = y f i y y f i mgdy = mgy f + mgy i = U f + U i Since U depends only on y, any given height is a plane of constant potential enegy: an equipotential. Enegy consevation: E total = K + U = = ½ mv 2 + mgy = const Examples: Inclined planes, pendulums, olle coastes, Mt. Lemmon, etc.

Gand Canyon Topo Map Bown contous ae lines of constant altitude: gavitational equipotentials. Numbes ae feet above sea level. Whee lines ae closely packed togethe, the slope is steep: the gadient is lage.

Gavitational equipotentials at Eath s suface. Discuss slopes (gadients), foces, wok, and path independence. Find Do A? B

Saddle points: thei contous, and thei peculia equilibium points

Electic potential enegy in a unifom electic field The electic foce field is unifom (and pointed downwad in this example) with a constant foce: F = -qe Wok-enegy theoem U = qey W y y f = Fdy = qedy i y y f i = qey f + qey i = U f + U i Since U depends only on y, any given height is a plane of constant potential enegy: an equipotential. Enegy consevation: E total = K + U = = ½ mv 2 + qey = const Examples: electon micoscope beams, linea acceleatos,

The electic foce is consevative As in the case of gavity, the wok done is path independent.

What, pecisely, is electic potential??? Electic potential is the electic potential enegy pe unit chage: V E = V = U q E 0 Units : J C = [ V] (volts) So a paticle o object with chage q at a location whee the electic potential is V will have a potential enegy U = qv. ΔU = qδv. Chaged plates again, with given voltages: 600 V ΔU qeδy ΔV V ΔV = = = Ed E = q q d m fom which we can find the electic field, E = (600V 100V)/(.2 m) = 2500 V/m =.2 m 100 V

Alessando Giuseppe Antonio Anastasio Volta 1745-1827 Invented electophous (chaging device), the battey (silve, zinc, and bine), discoveed methane, ignited gases with electic spaks, Volt / mete = Newton/Coulomb Volta demonstating his battey to Napoleone in 1801, by Giuseppe Betini

ΔV is easily ceated with batteies o powe supplies A chaged capacito

We can analyze poblems with dynamics o enegetics Example: An acceleato/deceleato. (Discuss.) Dynamics: F = ma F = qe = ma a = qe/m Then, we would do kinematic analysis. Consevation of total enegy: K + U 1 2 2 1 2 mv i + qvi = mv f + 2 qv This method can be applied to any system fo which we know V! f

Two ways to calculate V fo new systems 1. As we ve been doing, stat fom the wok-enegy theoem: W = ΔU = F ds ΔU = F ds ΔV = ΔU q 0 2. O, notice that if we divide both sides of the middle equation above by the test chage, we can find V diectly fom the electic field: ΔU q 0 = F ds q 0 ΔV = E ds In most cases, we ll find the second appoach moe convenient fo finding the electic potential.

Electic potential of a point chage This esult will be the building block used to calculate V systems of point chages o continuous chage distibutions: kq kq kq ΔV = Ed = d = = 0 = V V 2 We have chosen the potential to be zeo at infinity, giving us this vey simple esult to cay fowad: the electic potential of a point chage. V = kq No vecto quantities to complicate things!

But we also must be awae of the chage sign, and its effect on V

Since the electic foce is consevative, and the field is cental (along the line of centes between chages), the wok done on a chage moving between any two points is independent of the path taken. This allows fo the definition and calculation of an electic potential, V. Path independence again

The many epesentations of V fo a positive chage V = kq

Electic potential obeys the supeposition pinciple Fo any system of point chages V = V = i i i kq i i Fo this system of 3 chages: kq 1 2 V = + + 1 kq 2 kq 3 3 No vecto algeba to fuss with! But, in geneal we may need to calculate the distances using: i = ( x x ) + ( y y ) 2 + ( z z ) 2 i 2 i i Whee position a is (x,y,z), and the ith chage is at (x i,y i,z i ).

Equipotentials of familia chage distibutions

Equipotentials of a dipole two epesentations

Do Some potentially inteesting examples

Electic potential fo continuous chage distibutions We can find V fo continuous chage distibutions stating fom the equation fo a point chage. The method is simila to that used to find F o E, but without the need fo vecto quantities in the integand. point chage diffeential fom kq kdq V = dv = integate to find V V = k dq pick dq based on the given chage distibution dq = λ ds o dq = σda o dq = ρd 3 Tun the cank and the answe will appea

Electic potential of a unifom ing of chage We can avoid doing an integation hee! How? Wite down the answe immediately.

Electic potential of a disk of unifom chage Apply the same tactic we used fo the electic field of a disk: divide the disk into infinitesimal ings of potential dv, and integate fom 0 to R.

Compaing potentials of point chage and disk chage Sketch

Set up Electic potential of a finite line of chage

Electic potential of an infinite line o cylinde of chage This calculation is consideably easie than fo the finite line because we can use cylindical symmety, knowing that all electic field lines ae diected adially, pependicula to the line of chage. Integating the 1/ electic field ove, we get ln() dependence. V If we ty setting V=0 at infinity, we get nonsense since the chage itself extends to infinity. a Vb = Ed = V a b a 0 = 2kλ ln = a b a 2kλ d = 2kλ ln b a But we can still use the fist esult to calculate voltage diffeences, which is all we will need. V a V b = 2kλ ln b a

Showing that electic field lines ae pependicula to equipotentials E pependicula E Equipotentials V constant V 2 V 1 E paallel If thee wee a component of E paallel to the equipotential, it would cause a foce F =qe in that diection on any chage q. So it could do wok along the equipotential. But no wok is done on paticles moving along an equipotential, so E must be pependicula to the equipotential. Q.E.D.

Recall Conductos in electostatic equilibium: electic field popeties fom self-consistency I. In electostatic equilibium, thee is no field inside a conducto. Othewise, the chages inside would continue to move! (F = qe) II. Stating point: In conductos, chages ae fee to move, both inside and on the suface. When they have stopped moving, the excess chages ae in electostatic equilibium. In electostatic equilibium, all excess chages ae on the suface of a conducto. Othewise, thee would be electic fields inside! III. In electostatic equilibium, the electic field outside the conducto, and nea the suface, is pependicula to the suface. Othewise, excess chages would continue to move along the suface! So

Additional popeties of conductos Combining the infomation on the two pevious pages IV. In electostatic equilibium, the sufaces of conductos ae equipotentials. Because, at any point whee the electic field contacts the suface it must be pependicula to the conducto and to the local equipotential. The two must coincide. V. In electostatic equilibium, the inteio of a conducto is an equipotential volume at the same potential as its suface. This agument goes back to the popety that thee ae no electic fields inside conductos. So thee ae no foces in the inteio to do wok on chages thee. And since the inteio is continuous with the suface, its potential matches that of the suface. These popeties geatly simplify the sketching of electic fields and equipotentials in the pesence of conductos.

Electic field and potential of a spheical shell of unifom σ 1. Assume this is a conducto and discuss E and V in the inteio. 2. What would we see if this is an insulato? 3. Estimate the chage on the dome of ou Van de Gaff machine.

Equipotentials of chaged sphee nea an infinite plane 1. Do these look like insulatos, conductos, o a mix? 2. Discuss point chage nea a gounded conducting plane Method of images.

Finding the electic field fom the electic potential So fa, we have been using this equation to find the electic potential when the electic field is known. ΔV = E ds But what if we wee given the electic potential, and wished to find the electic field? (The invese opeation.) Let s look at this in 1D fist: Equation above in 1D So fo any poblem in 1D we can use: = Edx dv dx d = dx V ( E ) xdx = Ex E x Diffeentiate both sides w..t. x V = x E y V = y E z V = z Then, the 2D o 3D equation can be found by eassembling the vecto E: E = ie ˆ ˆ x + jey + kˆ V V ) V E E = iˆ + z + ˆj k x y z E = iˆ + ˆj + kˆ V = V E = V x x x E = - (the gadient of V) o o

Examples: electic field fom potential 1. Fo any example whee we have found the potential along some axis (ing of chage, disk of chage, line of chage, sphee of chage, ) we can use the following to find (o coss-check) the electic field along that same axis: V V V E x = o E y = o E x y z = z 2. Fo any poblem whee we ae given the potential as a function of x, y, and z, we can take (-) the gadient of that function in 1D, 2D, o 3D to find the the vecto E as a function of x, y, and z. E = V Eg: V 3 2 2 = 3x + 4xy 2z + 6x z 2 13

Paticles in collision. And bound systems such as atoms.

Othe potentially inteesting examples

Geige counte