-19Suppose an electron (q = -e = - 1.6 × 10 C,m=9.1×10¬-31kg) is accelerated from rest through a potential difference of Vab= +5000V. 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 Vabrelated by the equation:U =Assuming all potential energy U is converted to kinetic energy K,K + U = 0K= -U1Since K= mv² and using the formula for potential energy above, we arrive at an equation for speed:v = ( 2eV/m1/2Plugging in values, the value of the electron's speed is:v= 4.1x 107 m/s

Potential energy = Charge * Potential difference U = q Vab q = -e , Hence, U = - e × Vab K = - U…

Suppose an electron (q = -e= -1.6 × 10' 'C,m=9.1 × 10¯³ª 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: v = ( 2eV/m 1/2

Suppose an electron (q = -e= -1.6 × 10' 'C,m=9.1 × 10¯³ª 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: v = ( 2eV/m 1/2

Physics for Scientists and Engineers

10th Edition

ISBN:9781337553278

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter24: Electric Potential

Section: Chapter Questions

Problem 2P: (a) Find the electric potential difference Ve required to stop an electron (called a stopping...

See similar textbooks

Similar questions

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?
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?
An electron moving parallel to the x axis has an initial speed of 3.70 106 m/s at the origin. Its speed is reduced to 1.40 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at the higher potential?
(a) Find the electric potential difference Ve required to stop an electron (called a stopping potential) moving with an initial speed of 2.85 107 m/s. (b) Would a proton traveling at the same speed require a greater or lesser magnitude of electric potential difference? Explain. (c) Find a symbolic expression for the ratio of the proton stopping potential and the electron stopping potential. Vp/Ve.
For the arrangement described in Problem 26, calculate the electric potential at point B, which lies on the perpendicular bisector of the rod a distance b above the x axis. Figure P20.26
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?
(a) Find the potential difference VB required to stop an electron (called a slopping potential) moving with an initial speed of 2.85 107 m/s. (b) Would a proton traveling at the same speed require a greater or lesser magnitude potential difference? Explain. (c) Find a symbolic expression for the ratio of the proton stopping potential and the electron stopping potential, Vp/Ve. The answer should be in terms of the proton mass mp and electron mass me.
Four charged particles are at rest at the corners of a square (Fig. P26.14). The net charges are q1 = q2 = +2.65 C and q3 = q4 = 5.15 C. The distance between particle 1 and particle 3 is r13 = 1.75 cm. a. What is the electric potential energy of the four-particle system? b. If the particles are released from rest, what will happen to the system? In particular, what will happen to the systems kinetic energy?
The distance between two small charged spheres with charges qA = 8.35 C and qB = +4.90 C is 48.0 cm. a. What is the electric potential energy due to the two spheres? b. What is the electric potential halfway between the two spheres along the line connecting them?
(a) Calculate the electric potential 0.250 cm from ail electron, (b) What is the electric potential difference between two points that are 0.250 cm and 0.750 cm from an electron? (c) How would the answers change if the electron were replaced with a proton?
Rank the potential energies of the four systems of particles shown in Figure OQ20.6 from largest to smallest. Include equalities if appropriate. Figure OQ20.6
The three charged particles in Figure P25.22 are at the vertices of an isosceles triangle (where d = 2.00 cm). Taking q = 7.00 C, calculate the electric potential at point A, the midpoint of the base.