1Lecture 4 Electric Potentialand/ Potential Energy Ch. 25•Review from Lecture 3•Cartoon - There is an electric energy associated with the position of acharge.•Opening Demo -•Warm-up problems•Physlet•Topics•Electric potential energy and electric potential•Calculation of potential from field•Potential from a point charge•Potential due to a group of point charges, electric dipole•Potential due to continuous charged distributions•Calculating the filed from the potential•Electric potential energy from a system of point charge•Equipotential Surface•Potential of a charged isolated conductor•Demos•teflon and silk•Charge Tester, non-spherical conductor, compare charge density atRadii•Van de Graaff generator with pointed objects2345Potential Energy and Electric potential• The electric force is mathematically the same as gravity so it too must be aconservative force. We will find it useful to define a potential energy as is thecase for gravity. Recall that the change in the potential energy in moving fromone point a to point b is the negative of the work done by the electric force.• = - W = -Work done by the electric force =• Since , ! U = and• Electric Potential difference = Potential energy change/ unit charge• (independent of path, ds)!q0E " dsab#!"#badsFabUUU !="F = q0E!V =!Uq0!"#=#=$ dsEVVVabSI unit of electric potential is volt (V):1 Volt = 1 Joule/Coulomb (1 V = 1 J/C)• Joule is too large a unit of energy when working at the atomic or molecular level,so use the electron-volt (eV), the energy obtained when an electron moves through a potential difference of 1 V. 1 eV = 1.6 x 10-19 J6 (independent of path, ds)Therefore, electric force is a conservative force. = - Work done by the electric force = ! F " dsif#!U = Uf" UiqUV!=!!"#=#=$ dsEVVVifyx7•The potential difference is the negative of the work done per unit charge by an electric field on a positive unitcharge when it moves from one point to another.• V is a scalar not a vector. Simplifies solving problems.•We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at the negative terminal of a battery or0 at infinity for a point charge. !V ="Wq0= "!Fq0# d!s$= "!E # d!s$8Example of finding the potential difference ina Uniform FieldWhat is the electric potential difference for a unit positive charge moving in anuniform electric field from a to b?EEdabx direction)( abbabaxxEdxEdsEV !!=!="!=#$$dEdV !="VqU !=!qEdU !="dV = !EdxE = !dV / dx9Example for a battery in a circuit• In a 9 volt battery, typically used in IC circuits, the positive terminal has apotential 9 v higher than the negative terminal. If one micro-Coulomb ofpositive charge flows through an external circuit from the positive to negativeterminal, how much has its potential energy been changed?qVVVqUV ab )90( !=!="="Potential energy is lower byqU 9!="9 µJCV6101)9( !""!=JoulesU6109!"!=#!U = " 9 microJoules = " 9 µJWe also assumed that the potential at b was 010Example of a proton accelerated in a uniformfieldA proton is placed in an electric field of E=105 V/m and released. After going 10cm, what is its speed?Use conservation of energy.a b+E = 105 V/md = 10 cmmqEdv2=v =2 ! 1.6 ! 10"19C ! 105Vm! 0.1m1.67 ! 10"27kgv = 1.4 ! 108msEdVVVab!=!="!K = qEdUK !"=!!U = q!V = "qEd0=!+! KU12mv2= qEd11What is the electric potential when movingfrom one point to another in a field due to apoint charge? !V = "!E # d!r$ !E =kqr2ˆr Vf! Vi= !!E " d!rif#12 Vf! Vi= !!E " dˆrR#$= !kq cos 0"1r2dR#$r = kq1rR#= kq (1#!1R)RkqV =041!"=keqn 25-26Replace R with rV =14!"0qrVf! Vi= 0 ! Vi= !kqRPotential of a point charge at a distance R Vf! Vi= !!E " dˆrif#13Electric potential for a positive point chargeV (r) =kqrr = x2+ y2• V is a scalar• V is positive for positive charges, negative for negative charges.• r is always positive.• For many point charges, the potential at a point in space is thesimple algebraic sum (Not a vector sum)14Hydrogen atom.• What is the electric potential at a distance of 0.529 A fromthe proton? 1A= 10-10 mElectric potential due to a positive point charger = 0.529 AmCCNmRkqV101922910529.106.11099.8!!"""#$%&'("==V = 27.2JC= 27.2VoltsWhat is the electric potential energy of the electron at that point?U = qV= (-1.6 x 10-19 C) (27.2 V)= - 43.52 x 10-19 Jor - 27.2 eV where eV stands for electron volts.Total energy of the electron in the ground state of hydrogen is - 13.6 eVAlso U= 2E = -27.2 eV. This agrees with above formula.15What is the electric potential due to severalpoint charges?• For many point charges, the potential at a point in space is the simple algebraic sum(Not a vector sum)!"#$%&++=332211rqrqrqkVV =kqirii!r1xr3yr2q1q2q31617Potential due to a dipoleFor two point charges, the total potential is the sum ofthe potentials of each point charge.batotaldipoleVVVV ,So +==!"#$%&'+=+=babadipoler)q(rqkVVV!"#$%&'=baabrrrrkqWe are interested in the regime where r>>d. Vdipole! kqd cos!r2!kp cos!r2where p is the dipole moment.As in fig 2, ra and rb are nearly parallel. And thedifference in their length is dcos". Also because r>>d,ra rb is approximately r2.1819Potential due to a ring of charge• Direct integration. Since V is a scalar, it is easier to evaluate V than E.• Find V on the axis of a ring of total charge Q. Use the formula for a pointcharge, but replace q with elemental charge dq and integrate.Point chargeFor an element of charger is a constant as we integrate.This is simpler than finding E because Vis not a vector.rkqV =rkdqdVdq = ,!=rkdqVQ)Rz(kV22+=!!+= dqRzk)(22!+=)(22Rzkdq20Potential due to a line chargerdqkdV =We know that for an element of charge dqthe potential isrdxkVd ,So!=For the line charge let the charge density be #.Then dq=#dx22dxr ,But +=22dxdxkVd ,Then+!=Now, we can find the total potential V produced by the rod at point P byintegrating along the length of the rod from x=0 to x=L!!!+"=+"==L022L022L0dxdxkdxdxk VdV! V = k"ln(x + x2+ d2)0L)dln)dLL(ln(k Vo,S22!++"=Or, V = k!lnL + L2+ d2d"#$%&'21A new method to find E if the potential is known. If we know V, how do we find E?So the x component of E is the derivative of V with respect to x, etc.–If V = a constant, then Ex = 0. The lines or surfaces on which Vremains constant are called equipotential lines or surfaces.–See example on next slide !V = "!E # d!s$ dV = !!E " d!sdzdVEdydVEdxdVEzyx!=!=!=dzEdyEdxEdVzyx!!!= d!s =ˆidx +ˆjdy +ˆkdz !E = Exˆi +