9. 1 points SerPSET9 25 P005. My Notes A uniform electric field of magnitude 285
ID: 1871487 • Letter: 9
Question
9. 1 points SerPSET9 25 P005. My Notes A uniform electric field of magnitude 285 V/m is directed in the negative y direction as shown in the figure below. The coordinates of point are (-0.400, -0.85o) m, and those of pointare (0.700, 0.900) m. Calculate the electric potential difference VB-VA using the dashed-line path Need Help? Read Weich 10. 1 points SerPSET9 25 P009 My Notes A particle having charge q-+2.40pc and mass 0.0100 k is connected to a string that is 1.60 m long and tied to the pivot point P in the figure below. The particle, string, and pivot point all lie on a frictionless, horizontal table. The particle is released from rest when the string makes an angle . 60.0 with a uniform electric field of magnitude E-260 v/m. Determine the speed of the particle when the string is parallel to the electric field. m/s Top viewExplanation / Answer
9)
From A to C(-0.400, 0.900) m,
VC - VA = E*[0.900 - (-0.850)]
VC - VA = 285*1.75
VC - VA = 498.75 V
From C to B, VB - VC = 0
VB - VA = VC - VA = 498.75 V
10)
at Equillibirum point ,
Solving it using conservation of energy
Vf^2 = (2qEL(1 - cos(theta)) / m
Vf^2 = (2 x 2.4 x 10^-6 x 260 x 1.60 x (1 - cos(60)) / 0.0100
Vf = 0.316 m/s
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