Considering electron and proton as two charged particles separated by d = 5.9 x 10-11 m calculate the potential energy of the electron in the field of the proton. Take the mass of the proton 1.7 × 10-27 kg, the mass of the electron 9.1 × 10-31 kg, the electron charge –1.6 × 10-19 1 = 9 x 10° m/F. Give your answer in atto-Joules (aJ, 10-18 J). 4πεο C and

Considering electron and proton as two charged particles separated by d = 5.9 x 10-11 m calculate the potential energy of the electron in the field of the proton. Take the mass of the proton 1.7 × 10-27 kg, the mass of the electron 9.1 × 10-31 kg, the electron charge –1.6 × 10-19 1 = 9 x 10° m/F. Give your answer in atto-Joules (aJ, 10-18 J). 4πεο C and

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...

See similar textbooks

Similar questions

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)?
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?
Suppose an electron (q = - e = - 1.6 x 10¬19 C,m=9.1 × 10¬3' 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 mv and using the formula for potential energy above, we arrive at an equation for speed: 2 Since K= 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 × 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=
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= - 1.6 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=-1.6x10^-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:
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
The electric field in a certain region is given by the equation E=(axn − b)i, where a = 12 N/(C⋅mn), b = 5 N/C and n = 6. Calculate the electric potential difference (ΔV = V2 − V1), in volts, between the points x2 = 1.6 m and x1 = 0.65 m, i.e., ΔV = V(x2) − V(x1). ΔV =
Calculate the speed of an electron (in m/s) after it accelerates from rest through a potential difference of 140 V. 7.02e25 m/s The speed of the electron v can be found using the kinetic energy equation, K = 1/2(mv^2). The change in kinetic energy is equal to the work done on the electron, K = W = -U = -qV. Using those two equations, solve for the speed v in terms of the potential V and constants. Remember that the mass of an electron is 9.1e-31 kg and its charge is 1.6e-19 C. Submit Answer Incorrect. Tries 1/2 Previous Tries
Considering electron and proton as two charged particles separated by d = 4.5 × 10-11 m calculate the electrostatic potential at a distance d from the proton. Take the mass of the proton 1.7 x 10-27 kg, the mass of the electron 9.1 x 10-31 kg, the electron charge -1.6 × 10-19 C and 1 Απερ = 9 × 10⁹ m/F.
In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV and shot onto a phosphorescent screen to produce an image. What is the speed of the electrons when they reach the screen? (g. = 1.602 x 10-19C ;me = 9.11 x 10 -31 kg) Answer: x10' m (express your answers in tenths place or one decimal digit only)