1. A straight wire AB of length 40 cm is charged 20 µC. Both ends wires A and B are located at coordinates A (-40.0) and B (0.0). Point P is at 30 cm above point B (0.30). Determine: a. Electric field at P due to charged straight wire. b. Electric force experienced by Q = +10-7C when it is located at P. 30 cm 40 cm
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- A negatively charged particle, q1, is brought near a stationary, negatively charged particle, q2. The charge, q1, is then allowed to move freely, while q2 remains stationary. Which of the following statements is true for the moving charge? A). ΔU > 0; ΔV> 0 B). ΔU < 0; ΔV< 0 C). ΔU > 0; ΔV< 0 D). ΔU < 0; ΔV> 0 E). ΔU = 0; ΔV< 03. A charged particle experiences a force of 150 N when placed in an electric field of magnitude ɛ. If the same charged particle is placed in an electric field of magnitude 5ɛ, determine the resulting force. a. 750 N b. 30 N c. 300 N d. 75 N e. none of the above3 The figure represents two charged parallel plates with 0.2m distance between them. How much the electric field 1200 +. +. +. V. (d) (b) (a) Y 6000N/c 240N/c 60N/C 24x10 N/c Acti vate Wi Go to Sen NEXE + Previous delets prt sc
- 1-32. Suppose we have an infinitely long wire with uniform line charge density λ = -1.00 × 10-⁹ C/m lying flat on the ground. The wire is fully stretch out (not coiled up). a) How strong is the repulsive electric force on an electron placed 1.00 m above the wire? Hint: this is another Gauss' Law problem, just with some extra calculations after. b) How does this compare the gravitational force on the electron pulling it down?1) An electron is released from rest in a weak electric field given by -2.9 x 10-10 N/C Ĵ. After the electron has traveled a vertical distance of 1.9 µm, what is its speed? (Do not neglect the gravitational force on the electron.) mm/s Submit You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question.
- How do you rearrange the electric force equation to distance? A. d=Fe/kq1q2 B. d= square root kq1q2/Fe C. OtherTwo positive charges, each of magnitude q, are on the y-axis at points y= ta and y=-a Where would a third positive charge of the same magnitude be located for the net force on the third charge to be zero? a. at the origin b. at y= 2a c. at y=-2a d. at y= -a9 m +2 C +7C Determine the electric force exerted on sphere A. 1.4 1010 newtons 1.3 1011 newtons 9'10 newtons 1.6 10 newtons
- A proton is projected in the positive x direction into a region of uniform electric field E = (-5.30 x 105) i N/C at t = 0. The proton travels 6.40 cm as it comes to rest. (a) Determine the acceleration of the proton. 0.9638e7 X magnitude How..do.you find the acceleration of an object if you know the net force that acts on it? m/s² direction --Select- (b) Determine the initial speed of the proton. 0.351e7 X magnitude The electric field is constant, so the force is constant, which means the acceleration will be constant. m/s direction -Select-- (c) Determine the time interval over which the proton comes to rest. 0.3640 You appear to have calculated the time correctly using your incorrect results from parts (a) and (b). sIn a homogeneous electric field, an electron e and a proton p are simultaneously liberated from rest, as depicted. Assume that the distance between the particles is adequate for the electric field to be the sole force acting on each particle after release. When the particles are still in the field at a later moment, the electron and proton will have the same a. direction of motion b. velocity c. displacement d. magnitude of acceleration e. magnitude of force acting on them.6| 17 aölJI - 2 uacl X 10 uluilg (-) X Content Google dai X 9 https://blackboard.uob.edu.bh/ultra/courses/ 24065 1/cl/outline Not syncing Remaining Time: 29 minutes, 53 seconds. * Question Completion Status: Consider a uniformly charged ring of radius R-0.2 m and linear charge density 2-3 µC/m as shown in the figure. Find the electric potential (in V) at (p) a distance 0.7 m from the center. 0 0.7 m 3.1x 10* 6.2 x 104 O9.3x 104 7.8x 104 O 4.7x10 7:05 PN 4/3/202 arch