Chapter 20. Electric Potential Electric Potential Energy
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1 Chapter 20 Electric Potential Electric Potential Energy
2 CONSERVTIVE FORCES conservative force gives back work that has been done against it Gravitational and electrostatic forces are conservative Friction is NOT a conservative force
3 POTENTIL ENERGY The change ΔU B in potential energy, associated with a conservative force, is the negative of the work done by that force, as it acts from point to point B ΔU B = -W B ΔU B = U B U = potential energy difference between and B
4 POTENTIL ENERGY IN CONSTNT FIELD E E L B The potential energy difference between and B equals the negative of the work done by the field as the charge q is moved from to B ΔU B = U B U = -W B = -F E L = q E L ΔU B =q E L
5 POTENTIL ENERGY IN CONSTNT FIELD E The potential energy difference between and B equals the negative of the work done by the field as the charge q is moved from to B ΔU B = U B U = - F E L B ΔU B = q E L when the +q charge is moved against the field
6 POTENTIL ENERGY ΔU B = U B U = -W B = -F E L Potential energy is a relative quantity, that means, it is always the difference between two values, or it is measured with respect to a reference point (usually infinity). We will always refer to, or imply, the change in potential energy (potential energy difference) between two points.
7 ELECTRIC POTENTIL DIFFERENCE The potential energy ΔU depends on the charge being moved. In order to remove this dependence, we introduce the concept of electric potential ΔV Electric Potential = Potential Energy per Unit Charge ΔV B = ΔU B / q ΔV B = V B V Electric potential difference between the points and B
8 ELECTRIC POTENTIL IN CONSTNT FIELD E E L B The electric potential difference between and B equals the negative of the work per unit charge, done by the field, as the charge q is moved from to B ΔV B = V B V = -W B /q = qe L/q = E L ΔV B = E L
9 POTENTIL ENERGY IN CONSTNT FIELD E ΔU B E L ELECTRIC POTENTIL IN CONSTNT FIELD E ΔV B B ΔV B = ΔU B / q ΔU B = U B U = -W B = -F E L ΔV B = V B V = -W B /q = E L ΔU B =q E L ΔV B = E L
10 UNITS Potential Energy ΔU: [Joule] [N m] (energy = work = force x distance) Electric Potential ΔV: [Joule/Coulomb] [Volt] (potential = energy/charge) Electric Field E: [N/C] [V/m] (electric field = force/charge = potential/distance)
11 Relation Between Electric Field and Electric Potential E B Since the field E is constant ΔV B = E L Then, at a distance X from plate X L E L ΔV X = E X In general: ΔV E = or Δ V = EΔx Δx
12 For the parallel plate capacitor shown below make a graph (Y vs X) of: a) The electric field as a function of distance (inside the plates) b) The electric potential as a function of distance (inside the plates) E V x x
13 CONSERVTION OF ENERGY Since the electric force is conservative, total energy is conserved in electrostatics For a charged object in an electric field: K initial + U initial = K final + U final Expressing the kinetic energy as K = ½ mv 2, the potential energy as U = qv, and assuming movement from to B 1 mv + qv = 1 mv + qv B B
14 1 mv + qv = 1 mv + qv B B 1 mv 1 mv = qv qv B B n increase in kinetic energy corresponds to a decrease in potential energy E 1 2 Where is the potential larger, at 1 or at 2?
15 Motion of a Point Charge in an Electric Field e + + B If the potential difference between and B is +1 V, then, a particle of charge +e, released at, will gain a kinetic energy of 1 electron volt [1eV] as it reaches B E K = ev ev B = e ΔV B if ΔV B = 1 V K = 1 ev (electron volt) 1 ev = 1.6 x J ELECTRON VOLT
16 Motion of a Point Charge in an Electric Field particle of mass m and charge q, placed in an electric field E, will experience a force F = qe F = q E = m a q a E The particle will accelerate with acceleration: a = (q/m) E In one dimension the motion of the particle is described by: x = x 0 + v 0 t + ½ a t 2 v = v 0 + a t v 2 = v a (x x 0 )
17 Motion of a Point Charge in an Electric Field Between the plates of a parallel plate capacitor (in vacuum) q m L V a The charge accelerates with a = (q/m) E = (q/m) (V/L)
18 Electric Potential of Point Charges If a test charge q 0 is released at, it will gain kinetic energy, and lose potential energy, as it moves towards B. The change in potential energy is: U kq0q kq0q UB = r r B The corresponding change in electric potential (for charge q) is: V kq VB = r kq r B
19 Electric Potential of Point Charges V kq VB = r kq r B If r B and V, thenv 0 kq = r B Electric Potential for a Point Charge Potential Energy of a Point Charge kq V = r unit V volt : ( ) kq0q U = q0 V = unit : J Joule r ( )
20 Superposition The total electric potential due to two or more charges is equal to the algebraic sum of the potentials due to each charge separately V = kq r The electric potential energy of a group of charges is equal to the sum of the potential energies for each pair of charges U kqiq = r ij j
21 Equipotential Lines (Surfaces) E B Since the field E is constant ΔV B = E L X L E Then, at a distance X from plate ΔV X = E X L ll the points along the dashed line, at X, are at the same potential. The dashed line is an equipotential line
22 Equipotential Lines (Surfaces) X E L The electric field is always perpendicular to the equipotential lines It takes no work to move a charge at right angles to an electric field The electric field points in the direction in which the electric potential decreases EQUIPOTENTIL ELECTRIC FIELD
23 Equipotential Lines (Surfaces) Electric Field E = kq r 2 Electric Potential V = kq r
24
25 Ideal Conductors Charge can be moved from one location to another, in an ideal conductor, with no work being done. Since the work is zero, the change in potential is zero. Every point is at the same potential Ideal conductors are equipotential objects lso, extra charge is distributed over the surface of a conductor
26 Ideal Conductors t the surface of a charged conducting sphere 2 ( 4πR ) kq kσ V = = = 4π kσ R R R 2 ( 4πR ) kq kσ E = = = 4π kσ 2 2 R R For two spheres at the same potential The sphere with radius R/2 must have twice the charge density, and therefore twice the electric field The charge density is more concentrated and the electric field is higher where a conductor is more sharply curved Note the field perpendicular to the surface
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AP Physics C Coulomb s Law Free Response Problems 1. Four equal and positive charges +q are arranged as shown on figure 1. a. Calculate the net electric field at the center of square. b. Calculate the
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