Suppose an electron (q = – e= -1.6 × 10¬19 c,m=9.1 × 10¬31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K + U = 0 K = -U 1 -mv² and using the formula for potential energy above, we arrive at an equation for speed: Since K=

Suppose an electron (q = – e= -1.6 × 10¬19 c,m=9.1 × 10¬31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K + U = 0 K = -U 1 -mv² and using the formula for potential energy above, we arrive at an equation for speed: Since K=

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: Electric Potential And Capacitance

Section: Chapter Questions

Problem 16OQ: A filament running along the x axis from the origin to x = 80.0 cm carries electric charge with...

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A filament running along the x axis from the origin to x = 80.0 cm carries electric charge with uniform density. At the point P with coordinates (x = 80.0 cm, y = 80.0 cm), this filament creates electric potential 100 V. Now we add another filament along the y axis, running from the origin to y = 80.0 cm, carrying the same amount of charge with the same uniform density. At the same point P, is the electric potential created by the pair of filaments (a) greater than 200 V, (b) 200 V, (c) 100 V, (d) between 0 and 200 V, or (e) 0?
At a certain distance from a charged particle, the magnitude of the electric field is 500 V/m and the electric potential is 3.00 kV. (a) What is the distance to the particle? (b) What is the magnitude of the charge?
An electric potential exists in a region of space such that V = 8x4 2y2 + 9z3 and V is in units of volts, when x, y, and z are in meters. a. Find an expression for the electric field as a function of position. b. What is the electric field at (2.0 m, 4.5 m, 2.0 m)?
Suppose an electron (q = -e = -1.6 x 10¬9 C,m=9.1 x 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for %3D the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 Since K mv and using the formula for potential energy above, we arrive at an equation for speed: 2 v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=
Suppose an electron (q= -e= -1.6 x 10-19 C,m=9.1x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K +U = 0 K = -U Since K- and using the formula for potential energy above, we arrive at an equation for speed: v = ( 51/2 Plugging in values, the value of the electron's speed is: V= x 107 m/s
Suppose an electron (q= -e= -1.6x 10-19 C.m=9.1×10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 v. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U- Assuming all potential energy U is converted to kinetic energy K. K+U=0 K--U Since K=mv and using the formula for potential energy above, we arrive at an equation for speed: 1/2 Plugging in values, the value of the electron's speed is: x 10' m/s
Suppose an electron (q = - e= - 1.6 x 10-19 C.m = 9.1 x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U= 0 K= -U 1 Since K=mv and using the formula for potential energy above, we arrive at an equation for speed: 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=
Suppose an electron (q = -e= - 1.6 × 10¬19 C,m = 9.1 x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 mv and using the formula for potential energy above, we arrive at an equation for speed: Since K= V = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=
A) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of 0.965 cc? Express your answer in volts. B) What is the kinetic energy of the electron at this speed? Express your answer in joules. Express your answer in joules. C) What is the kinetic energy of the electron at this speed? Express your answer in electronvolts. Express your answer in electronvolts.
Consider 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.
Suppose an electron (q=-e=-1.6x10^-19 C, m=9.1x10^-31kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures.The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation:
What is the kinetic energy of an electron, in the unit of J, when it is accelerated by the electrical potential difference of 30.3 kV? Assume that the electron is initially at rest. Answer the value that goes into the blank. Be careful with units. ______ × 10-15 J