Transcript Slide 1
Chapter 17 Electrical Energy and Current 17 – 1 Electric Potential • Electric potential energy (EPE): the potential energy associated with an object’s position in an electric field • Electrical potential energy is a component of mechanical energy. ME = KE + PEgrav + PEelastic + PEelectric Electrical potential energy can be associated with a charge in a uniform field. ΔPEelectric = qEd magnitude of change in PE q = charge E = electric field d = displacement from a reference point Review • In the case of gravity, the gravitational field does work on a mass • WAB=F d = mgd • ΔPE = W AB= mgΔh • In the case of a charge moving in an electric field • As the positive charge moves from A to B, work is done • WAB = F d = q E d • ΔPE = W AB= q E Δd Energy and Charge Movements Direction of movement + charge - charge Along E Loses PE Gains PE Opposite E Gains PE Loses PE Electric potential energy for a pair of point charges q1q2 PE kc r Note: The reference point for zero potential energy is usually at Example • What is the potential energy of the charge configuration shown? q 3 = 1.0 mc q 2 = - 2.0 mc q 1 = 2.0 mc 0.30 m Given: q 1 = 2.0 x 10-6 C q 2 = - 2.0 x 10-6 C q 3 = 1.0 x 10-6 C r = 0.30 m (same for all) k= 9.0 x 109 n•m2/C2 The total potential energy is the algebraic sum of the mutual potential energies of all pairs of charges. • PET = PE12 + PE23 + PE 13 • = k q1q 2 + kq2q3 + kq1q3 r r • = k [(q1q2) + (q2q3 ) + r (q1q3)] r = [(9.0 x 109 n•m2/C2)/ 0.30] x [(2.0 x 10-6 C)(- 2.0 x 10-6 C) + (- 2.0 x 10-6 C)(1.0 x 10-6 C) +(2.0 x 10-6 C)(1.0 x 10-6 C)] =- 0.12 J Electric Potential • Electric potential: the electric potential energy associated with a charged particle divided by the charge of the particle • symbol for electric potential = V V = PE/q • SI unit = volt (V) • 1Volt = 1 Joule/Coulomb Potential Difference • Potential Difference equals the work that must be performed against electric forces to move a charge between the two points in question, divided by the charge. • Potential difference is a change in electric potential. PEelectric V q changein electric potentialenergy potentialdifference electric charge Chapter 17 Potential Difference, continued • The potential difference in a uniform field varies with the displacement from a reference point. ∆V = –Ed Potential difference in a uniform electric field E = electric field d = displacement in the field Sample Problem • A proton moves from rest in an electric field of 8.0104 V/m along the +x axis for 50 cm. Find • a) the change in in the electric potential, • b) the change in the electrical potential energy, and • c) the speed after it has moved 50 cm. • a) V = -Ed = -(8.0104 V/m)(0.50 m) = -4.0104 V b) PE = q V = (1.610-19 C)(-4.0 104 V) = -6.4 10-15 J • C) Ei = Ef • KEi+PEi = KEf + PEf, since KEi=0 KEf = PEi – PEf = -PE 1/2 mpv2 = -PE v = 2 PE/m = 2(6.4x10-15 J)/1.67x10-27 kg)=2.8x106 m/s Chapter 17Difference, Potential point charges • At right, the electric potential at point A depends on the charge at point B and the distance r. • An electric potential exists at some point in an electric field regardless of whether there is a charge at that point. Chapter 17 • The reference point for potential difference near a point charge is often at infinity. • Potential Difference Between a Point at Infinity and a Point Near a Point Charge q V k c r Electric Potential of Multiple Point Charges- Superposition • The total electric potential at point “a” is the algebraic sum of the electric potentials due to the individual charges. Problem Solving with Electric Potential (Point Charges) • Remember that potential is a scalar quantity – So no components to worry about • Use the superposition principle when you have multiple charges – Take the algebraic sum • Keep track of sign – The potential is positive if the charge is positive and negative if the charge is negative • Use the basic equation V = kcq/r Example: Finding the Electric Potential at Point P 5.0 mC = = -2.0 mC 6 5 . 0 10 C V1 (8.99 10 9 Nm 2 / C 2 ) 1.12 10 4 V, 4.0m 6 ( 2 . 0 10 C) V2 (8.99 10 9 Nm 2 / C 2 ) 3.60 10 3 V (3.0m) 2 (4.0m) 2 Superposition: Vp=V1+V2 Vp=1.12104 V+(-3.60103 V) =7.6103 V