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Electric Potential

Electric Potential. Physics 102 Professor Lee Carkner Lecture 11. PAL #10. q 2 = +4 m C. Find E 1 and E 2 E 1 = 8.99X10 9 (2X10 -6 )/(3 2 ) = 1124 N /C r 2 2 = 3 2 + 4 2 r 2 = 5 m E 2 = 8.99X10 9 (4X10 -6 )/(5 2 ) = 1438 N /C Find q 1 and q 2 q 1 = 0 (right on x-axis)

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Electric Potential

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  1. Electric Potential Physics 102 Professor Lee Carkner Lecture 11

  2. PAL #10 q2 = +4 mC • Find E1 and E2 • E1 = 8.99X109(2X10-6)/(32) = 1124N /C • r22 = 32 + 42 • r2 = 5 m • E2 = 8.99X109(4X10-6)/(52) = 1438N /C • Find q1 and q2 • q1 = 0 (right on x-axis) • Can get q2 from triangle • tan q2 = ¾, q2 = 37 degrees • q2 is below X axis so is -37 r2 =5 m 3 m q2 E1 q2 4 m E2 q1 = +2 mC

  3. PAL #10 q2 = +4 mC • Find X and Y components • E1x = E1 cos q1 = 1124 cos 0 = 1124 N/C • E1y = E1 sin q1 = 1124 sin 0 = 0 • E2x = E2 cos q2 = 1438 cos -37 = 1148 N/C • E2y = E2 sin q2 = 1438 cos -37 = -865 N/C • Find total E vector and angle • Etotal,x = E1x + E2x = 1124 + 1148 = 2272 N/C • Etotal,y = E1y + E2y = 0 – 865 = -865 N/C • E2total = E2total,x + E2total,y • E = (22722 +8652)½ • E = 2431 N/C • tan qtotal = Etotal,x / Etotal,y = -865 / 2272 • qtotal = -20 degrees • 20 degrees below + X axis r2 =5 m 3 m E1 q2 q2 4 m E2 q1 = +2 mC

  4. Electrical Potential Energy • The value of the potential tells us how much potential energy is at that point per unit charge • V = PE/q or PE = Vq • 1V = 1 J/C • Which could be converted into work or kinetic energy

  5. Point Charges and Potential • Consider a point charge Q, what is the potential for the area around it? • At infinity the potential is zero • It can be shown that: V = ke Q / r • Q is charge providing potential, q is charge experiencing potential

  6. Finding Potential • If the potential is provided by an arrangement of charges: • Total V = V1 + V2 + V3 … • DV = -Ed • We don’t know the potential at anyone point, but we can find the difference between two points separated by a distance d

  7. Potential and Energy • The change in energy is: DPE = Vfq – Viq = q(Vf-Vi) = qDV DPE = -W • Energy could be converted into kinetic energy: DPE = DKE KEi + PEi = KEf + PEf • Particle could speed up or slow down

  8. Problem Solving • Particle moving between two points of different potential • Find DV and then DPE (=qDV) • Add or subtract DPE to initial energy of particle

  9. Signs • As a positive charge moves with the electric field, the particle gains kinetic energy and loses potential energy • Like dropping a ball in a gravitational field • the electric field does positive work • If a positive charge is forced backwards against an electric field, the particle loses kinetic energy and gains potential energy • like rolling a ball up a hill • the field “does” negative work

  10. Potential and Charges • A negative particle: • Opposite of positive particle • A negative particle has the most potential energy at low potential • The potential and the potential energy are two different things • Potential at a point is the same no mater what kind of test charge is put there

  11. High potential Low potential

  12. Work • Since energy must be conserved: • DPE + W = 0 or DPE = -W • Work done by the field is positive if it decreases the potential energy • The “natural” movement • When the charge is forced in the “unnatural” direction • The negative work done by the system is the positive work done on the system

  13. Down gain KE lose PE field does +work “natural” Up lose KE gain PE you do work field “does” negative work “forced” For negative particle, everything is backwards But high and low potential are still in the same place High potential + E Low potential

  14. Equipotentials Blue = field = E Dashed = potential = V • Each line represents one value of V • Particles moving along an equipotential do not gain or lose energy • Equipotentials cannot cross

  15. Next Time • Read Ch 17.7-17.9 • Homework, Ch 17: P 10, 20, 35, 46

  16. sign of DPE sign of DV sign of W naturally? + charge moves with E field + charge moves against E field -charge moves with E field -charge moves against E field Electric Potential Chart

  17. The above electric field, • increases to the right • increases to the left • increases up • increases down • is uniform

  18. Is it possible to have a zero electric field on a line connecting two positive charges? • Yes, at one point on the line • Yes, along the entire line • No, the electric field must always be greater than zero • No, but it would be possible for two negative charges • No, the electric field is only zero at large distances

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