College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- The figure below shows a charged particle, with a charge of q = +39.0 nC, that moves a distance of d = 0.190 m from point A to point B in the presence of a uniform electric field E of magnitude 255 N/C, pointing right. (a) What is the magnitude (in N) and direction of the electric force on the particle? (b) What is the work (in J) done on the particle by the electric force as it moves from A to B? (c) What is the change of the electric potential energy (in J) as the particle moves from A to B? (The system consists of the particle and all its surroundings.) PEB − PEA = (d) What is the potential difference (in V) between A and B? VB − VA =arrow_forwardPlease give me answer very fast in 5 min subarrow_forwardHELParrow_forward
- Topic: Electrical Potential A point charge q1 = +2.40 µC is held stationary at the origin. A second point charge q2 = -4.30 µC moves from the point x = 0.150 m, y = 0 m to the point x = 0.250 m. How much work is done by the electric force on q2?arrow_forwardConsider two points in an electric field. The potential at point 1, V1, is 29 V. The potential at point 2, V2, is 173 V. An electron at rest at point 1 is accelerated by the electric field to point 2. Write an equation for the change of electric potential energy ΔU of the electron in terms of the symbols given. Find the numerical value of the change of the electric potential energy in electron volts (eV). Express v2, the speed of the electron at point 2, in terms of ΔU, and the mass of the electron me.arrow_forward= Two point particles have equal charge q 1100 esu and different masses, m₁ = 25 gr and m₂ = 65 gr, are confined to move freely along the x axis. Note that esu is the unit of electric charge when working with units of mass in grams and distance in centimeters. = The electrical force between the particles is a repelling force which is derived from the potential energy function U(x1, x2) of the two-particles system. U(x1, x2) depends only on the distance of the particles and is given by U(x1, x2) - q2 | x2 − x1| At t = 0) the particle of mass m2 is at rest at the origin (x2 = 0), and the particle of mass m₁ begins to move from point x1 = 76 cm with velocity vo 134 cm/s towards particle m2. = a. What is the location of the center of mass (in cm) at t - 0? -21. b. What is the location of the center of mass (in cm) at t = 0.4 sec? -6.2 c. What will be the minimum distance (in cm) between the two particles? 6.46 X d. Once the distance between the particles is minimal, what will be the speed…arrow_forward
- The rectangle in the figure is 3 sidescapped by 5.0cm and 15cm, etq1 = −5μCetq2 = + 2.0μC. IfV = 0 at infinity, what is the electric potential (a) at corner A? (b) at corner B? (c) How much work does it take to move a third load q3 = +3.0 μC from point B to point A on the diagnonal of the rectangle? (d) Does this work increase or decrease the potential electrical energy of the three-load system? Is it greater than, less than or equivalent to the work required to move q3 on trajectories that are inside the rectangle but not on the diagnostic? (f) outside the rectangle? Please show formulas, thank youarrow_forwardso the distance is from -3 to 5 cm which is 8cm or .08m. I did parts a-e and need help for parts f-h. This is what I have so far: a: V1=1000V b: 3.2e^-16 J c. potential energy; no work needed to move particle as it leaves electric field d. 3.2e^-16 J e. m1<m2<m3<m4, v1>v2>v3>v4 f. distance =.08m -time takes particle 1 hit detector: .98e^-6 sec ---> .08m/.98e^-6 s= 8.16e^4 m/s -time takes particle 2 hit detector: 1.4e^-6 s --> .08m/1.4e^-6 s= 5.71e^4 m/s -time takes particle 3 hit detector: 1.7e^-6 s --> .08m/1.7^-6 s= 4.7053^4 m/s -time takes particle 4 hit detector: 1.96^-6 s --> .08m/1.96^-6 s= 4.082e^4 m/s I am not sure if i Made any mistakes so please let me know if i DO!arrow_forwardConsider a certain amount of a conducting liquid sprayed into 650 equal spherical drops. Each drop is charged to the same electric potential of 2.5 V (relative to the infinity where potential is zero). If all these 650 drops are combined into one large spherical drop, what is the electric potential of this large drop? The electric potential of the large drop, V = Units Select an answer ✓arrow_forward
- = A point charge Q1 = +5.8 µC is fixed in space, while a point charge Q2 = +2.8 nC, with mass 6.3 µg, is free to move around nearby. Calculate the electric potential energy of the system, in joules, when Q2 is located 0.42 m from Q₁. If Q2 is released from rest at a point 0.42 m from Q₁, what will be its speed, in meters per second, when it is 0.79 m from Q₁? m/sarrow_forwardConsider two points in an electric field. The potential at point 1, V1, is 31 V. The potential at point 2, V2, is 191 V. An electron at rest at point 1 is accelerated by the electric field to point 2. a) write an equation for the change of electric potential energy of the electron in terms of the symbols given. b) Express v2, the speed of the electron at point 2, in terms of electric potential energy, and the mass of the electron, me.arrow_forwardThe figure below shows a small, charged sphere, with a charge of q = +38.0 nC, that moves a distance of d = 0.171 m from point Ato point B in the presence of a uniform electric field E of magnitude 280 N/C, pointing right. (a) What is the magnitude (in N) and direction of the electric force on the sphere? magnitude n? (b) What is the work (in J) done on the sphere by the electric force as it moves from A to B? J?? (C) What is the change of the electric potential energy (in J) as the sphere moves from A to B? (The system consists of the sphere and all its surroundings.) PEB − PEA = ? J (D)What is the potential difference (in V) between A and B? VB − VA = ? Varrow_forward
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