GENERAL PHYSICS 2

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ACTIVITY 1

1 8 Electric Potential Energy in a Uniform Electric Field " Recall that for a conservative force Facting along the x-direction F = - AU or Ax FAX = -AU But FAx is equal to work Equation 2.1 WA-B = -AU -U. = UA - UB Where WAS is the work done from A to B, UA is the potential energy at point A, and Up is the potential energy at Us B Figure 2.1 a. test charge go in a uniform downward electric field E. The field exerts a downward force F. b. the field does positive work on q. as it moves from point A to point B Figure2-la shows a positive test charge q. in a uniform downward electric field E produced by two parallel plates. The field exerts a downward force F = q. Ed. Work is done by the electric force on the charge as it moves from point A to point B (figure 2-1b) W = Fd = q.Ed = UA - UB Note that if the test charge moves in the direction of the electric field, its potential energy decreases .On the other hand, if it moves against the direction for the electric field, the potential energy increases.. Using the work-energy theorem, W = AK. Therefore, UA - UR = AK UA - UB = KB - KA Rearranging terms; KATUA = KB + KB Thus, the conservation of mechanicalenergy is also applies. Example: A point charge of 3.0 ne (nanoCoulumb) with a mass of 4.0g is moved from x = 1.0 m to 1.5 m in an electric feld of 5.0 N/C with the same direction as the motion of the charge, a. How much work is done on the charge by the electric force? b. What is the change in the potential energy of the charge? c. Assuming that the charge started from rest, what is its speed at x = 1.5 m? Given: 3.0 nc = 3.0 X 10 m = 4.0 g = 40 x 10pkg E = 5.0 N/C d = 1.5m - 1.0 m == 0.5m Solution: C. Speed a. Work Using the work-energy W = Fd theorem, From Equation W= KEKo Therefore' W = qEd 1 W = (3.0 x 10-6)(3:010 -my? x 10-C) (0.5 m) W = 7.5 x 10 9N.m Since vo = 0 (assuming that the W = 7.5 x 10-9/ charge started from rest), 1 b. Change in Potential Energy W = -mv2 AU = -W = -7.5 x 10-9/ 7.5 x 10-9/ = = (4.0 x 10-3kg)v2 v = 19 x 10-3m/sProblem Solving (Answer without solution will not be considered) A point particle of charge 2.5 ne and mass 3.25 x10 3kg is in a uniform electric field directed to the right it is released from rest and moves to the right. After it has travelled 12.0 cm, its speed is 25 m/s. Find the a. Work done on the particle, b. Change in the electric potential energy of the particle