html5-img
1 / 23

Lecture 4 Electric Potential/Energy Chp. 25

Lecture 4 Electric Potential/Energy Chp. 25. Cartoon - There is an electric energy associated with the position of a charge. Opening Demo - Warm-up problem Physlet Topics Electric potential energy and electric potential Equipotential Surface Calculation of potential from field

kathleen
Download Presentation

Lecture 4 Electric Potential/Energy Chp. 25

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 4 Electric Potential/Energy Chp. 25 • Cartoon - There is an electric energy associated with the position of a charge. • Opening Demo - • Warm-up problem • Physlet • Topics • Electric potential energy and electric potential • Equipotential Surface • 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 • Potential of a charged isolated conductor • Demos • teflon and silk • Charge Tester, non-spherical conductor, compare charge density at Radii • Van de Graaff generator with pointed objects

  2. Electric potential • The electric force is mathematically the same as gravity so it too must be a conservative force. We will find it useful to define a potential energy as is the case for gravity. Recall that the change in the potential energy in moving from one point a to point b is the negative of the work done by the electric force. • U = Ub -Ua = - Work done by the electric force = • Since F=qE,  U = and • Electric Potential difference = Potential energy change/ unit charge •  V= =  U/q • V = Vb -Va = - E.ds (independent of path ds)

  3. U = Ub -Ua = - Work done by the electric force =  V= =  U/q V = Vf -Vi = - E.ds (independent of path ds) Therefore, electric force is a conservative force

  4. V = Vf -Vi = - E.ds (independent of path ds) • The SI units of V areJoules/ Coulomb or Volt. E is N/C or V/m. • The potential difference is the negative of the work done per unit charge by an electric field on a positive unit charge when it moves from point a to point b. • We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at infinity for a point charge.

  5. Examples • What is the electric potential difference for a positive charge moving in an uniform electric field? E E d x direction a b d V = -Ed U = q V U = -qEd

  6. Examples • In a 9 volt battery, typically used in IC circuits, the positive terminal has a potential 9 v higher than the negative terminal. If one micro-Coulomb of positive charge flows through an external circuit from the positive to negative terminal, how much has its potential energy been changed? q = -9V x 10-6 C U = -9V x 10-6 Joules U = -9 microJoules Potential energy is lower by 9 microJoules

  7. Examples A proton is placed in an electric field of E=105 V/m and released. After going 10 cm, what is its speed? Use conservation of energy. E = 105 V/m d = 10 cm a b + V = Vb – Va = -Ed U = q V U + K = 0 K = (1/2)mv2 K = -U (1/2)mv2 = -q V = +qEd

  8. Electric potential due to point charges • Electric field for a point charge is E = (kq/r2). The change in electric potential a distance dr away is dV = - E.dr. • Integrate to get V = -  kq/r2 dr = kq/r + constant. • We choose the constant so that V=0 at r = . • Then we have V= kq/r for point charge. • V is a scalar, not a vector • 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 the sum V=  kqi/ri. y 2 1 p r3 3 r1 r2 x

  9. Replace R with r eqn 25-26

  10. Electric potential for a positive point charge

  11. Electric potential due to a positive point charge Hydrogen atom. • What is the electric potential at a distance of 0.529 A from the proton? • V = kq/r = 8.99*1010 N m2//C2 *1.6*10-19 C/0.529*10-10m • V = 27. 2 J/C = 27. 2 Volts - r = 0.529 A r +

  12. Ways of Finding V • 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 point charge, but replace q with elemental charge dq and integrate. Point charge V = kq/r. For an element of charge dq, dV = kdq/r. r is a constant as we integrate. V =  kdq/r =  kdq/(z2+R2)1/2 =k/(z2+R2)1/2  dq = k/(z2+R2)1/2 Q This is simpler than finding E because V is not a vector. V = kq/r

  13. More Ways of finding V and E • Use Gauss’ Law to find E, then use V= -  E.ds to get V • Suppose Ey = 1000 V/m. What is V? V = -1000 y • More generally, If we know V, how do we find E? dV= - E. ds • ds = i dx +j dy+k dz and dV = - Exdx - Eydy - Ezdz Ex = - dV/dx, Ey = - dV/dy, Ez = - dV/dz. • So the x component of E is the derivative of V with respect to x, etc. • If Ex = 0, then V = constant in that direction. Then lines or surfaces • on which V remains constant are called equipotential lines or surfaces. • See examples (Uniform field) V = Ed

  14. Equipotential Surfaces • Example: For a point charge, find the equipotential surfaces. First draw the field lines, then find a surface perpendicular to these lines. See slide. • What is the equipotential surface for a uniform field in the y direction? See slide. • What is the obvious equipotential surface and equipotential volume for an arbitrary shaped charged conductor? • See physlet 9.3.2 Which equipotential surfaces fit the field lines?

  15. a) Uniform E field E = Ex , Ey = 0 , Ez = 0 Ex = dv/dx V = Ex d V = constant in y and z directions • Two point charges (ellipsoidal concentric shells) • Point charge (concentric shells)

  16. How does a conductor shield the interior from an exterior electric field? • Start out with a uniform electric field with no excess charge on conductor. Electrons on surface of conductor adjust so that: 1. E=0 inside conductor 2. Electric field lines are perpendicular to the surface. Suppose they weren’t? 3. Does E = s/e0just outside the conductor 4. Is s uniform over the surface? 5. Is the surface an equipotential? 6. If the surface had an excess charge, how would your answers change?

  17. V = constant on surface of conductor + + + + + + + + + + 1 Radius R 2 Dielectric Breakdown: Application of Gauss’s Law • If the electric field in a gas exceeds a certain value, the gas breaks down and you get a spark or lightning bolt if the gas is air. In dry air at STP, you get a spark when E = 3*106 V/m. To examine this we model the shape of a conductor with two different spheres at each end:

  18. V = constant on surface of conductor + + + + + + + + + + 1 Radius R 2 The surface is at the same potential everywhere, but charge density and electric fields are different. For a sphere, V= q/(4 0 r) and q = 4r2,then V = (/ 0 )*r. Since E = / 0near the surface of the conductor, we get V=E*r. Since V is a constant, E must vary as 1/r and  as 1/r. Hence, for surfaces where the radius is smaller, the electric field and charge will be larger. This explains why:

  19. V = constant on surface of conductor + + + + + + + + + + 1 Radius R 2 This explains why: • Sharp points on conductors have the highest electric fields and cause corona discharge or sparks. • Pick up the most charge with charge tester from the pointy regions of the non-spherical conductor. • Use non-spherical metal conductor charged with teflon rod. Show variation of charge across surface with charge tester. Cloud + + + + Van de Graaff Show lightning rod demo with Van de Graaff - - - -

  20. A metal slab is put in a uniform electric field of 106 N/C with the field perpendicular to both surfaces. • Draw the appropriate model for the problem. • Show how the charges are distributed on the conductor. • Draw the appropriate pill boxes. • What is the charge density on each face of the slab? • Apply Gauss’s Law.  E.dA = qin/0

  21. Slab of metal In a uniform Electric field E = 106 N/C - - - - - + + + + Gaussian Pill Box Slab of metal In a uniform Electric field EvaluateE.dA = qin/0 • Left side of slab - E*A + 0*A = A/0, E = - /0, = - 106 N/C *10-11 C2/Nm2 = -10-5 N/m2 .Right side of slab 0*A + E*A = A/0, E = /0, = 106 N/C *10-11 C2/Nm2 = +10-5 N/m2 (note that charges arranges themselves so that field inside is 0)

  22. What is the electric potential of a uniformly charged circular disk?

  23. Warm-up set 4 1. HRW6 25.TB.37. [119743] The equipotential surfaces associated with an isolated point charge are: concentric cylinders with the charge on the axis vertical planes horizontal planes radially outward from the charge concentric spheres centered at the charge 2. HRW6 25.TB.17. [119723] Two large parallel conducting plates are separated by a distance d, placed in a vacuum, and connected to a source of potential difference V. An oxygen ion, with charge 2e, starts from rest on the surface of one plate and accelerates to the other. If e denotes the magnitude of the electron charge, the final kinetic energy of this ion is: 2eV eVd Vd / e eV / d eV / 2 3. HRW6 25.TB.10. [119716] During a lightning discharge, 30 C of charge move through a potential difference of 1.0 x 108 V in 2.0 x 10-2 s. The energy released by this lightning bolt is: 1.5 x 1011 J 1500 J 3.0 x 109 J 3.3 x 106 J 6.0 x 107 J

More Related