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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

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Lecture 4 Electric Potential/Energy Chp. 25

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  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

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