Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0cm. Each plate has a surface charge density of 50.0 nC/m2. A proton is released from rest at the positive plate. Given information: Kinetic energy of proton when it reaches the negative plate = 9.94*10^(-17)J Potential difference between plates = 621.19V Magnitude of the electric field between the plates from the charge density = 5.647kN/C a) Determine the speed of the proton just before it strikes the
Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0cm. Each plate has a surface charge density of 50.0 nC/m2. A proton is released from rest at the positive plate. Given information: Kinetic energy of proton when it reaches the negative plate = 9.94*10^(-17)J Potential difference between plates = 621.19V Magnitude of the electric field between the plates from the charge density = 5.647kN/C a) Determine the speed of the proton just before it strikes the
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
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Chapter24: Gauss’s Law
Section: Chapter Questions
Problem 24.46P: A thin, square, conducting plate 50.0 cm on a side lies in the xy plane. A total charge of 4.00 108...
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Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0cm. Each plate has a surface charge density of 50.0 nC/m2. A proton is released from rest at the positive plate.
Given information:
Kinetic energy of proton when it reaches the negative plate = 9.94*10^(-17)J
Potential difference between plates = 621.19V
Magnitude of the electric field between the plates from the charge density = 5.647kN/C
a) Determine the speed of the proton just before it strikes the negative plate.
b) Determine the acceleration of the proton.
c) Determine the force on the proton.
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