An electron is released from rest at the negative plate of a parallel plate capacitor. The electric field between the plates points due east and has a magnitude of 8.0x10^4 N/C. The electron accelerates as it moves through the electric field toward the positive plate. By calculating the electric force on the electron and dividing by its mass, its acceleration is found to be 1.40x1016 m/s2. The same calculation for a proton in the same electric field gives an acceleration of 7.66x1012 m/s2, which is less than that of the electron due to the proton's greater mass.

1. 38. An electron is released from rest at the negative The electric field points due east and has a magnitude of
plate of a parallel plate capacitor. The charge per 8.0x 10^4 N/C. Determine the magnitude of the acceleration
unit area on each plate is o=1.8 x 10^-7 C/m^2, ofthe proton and the electron. Chck that your answers
and the plates are separated by a distance of 1.5 x areconsistent with part (c) of the Concept Questions.
10^-2m. How fast is the electron moving just before
it reaches the positive plate? a) Yes, both of them experience force from the electricfield,
magnitude = Eq
F = (8.99 x 10^9)(1.8 x 10^-7)(1.8 x 10^-7) / (1.5
x 10^-2)^2 = 1.3N b) The proton accelerates in the path ofthe electric field and
the electron follows the path opposite ofthe electric field.
Then...
Proton - East Electron - West
Ax = 1.3N / 9.11 x 10^-27 = 1.43 x 10^26 m/s c )F = ma ==> F/m = a
Although the forces are equal, the masses arenot.
The proton weighs more than an electron,therefore, the
proton's acc will be less than that of anelectron.
56. Two very small spheres are initially neutral and
separated by distance of 0.50 m. Suppose that 3.0 x 10 ^13 d) E = 8.0 x 104
electrons are removed from one sphere and placed on the q = 1.6 x 10-19
other. (a) What is the magnitude of the electrostatic force F = Eq = 8.0 x 104 x 1.6 x10-19 = 1.28 x 10-14
F/m = a
that acts on each sphere? (b) Is the force attractice or
Electron = 1.28 x 10-14/9.11 x10-31 = 1.40 x 1016m/s2
repulisve? Why?
Proton = 1.28 x 10-14/ 1.67x 10-27 = 7.66 x 1012m/s2
As we thought, the acc for electron isbigger.
A.) First we need to know what the charge of each sphere
18.The drawing shows three point charges
would be. Since 1 electron has an elementary charge of -
fixed in place. The charge at the
1.6x10^-19C, we can find the total charge of the electrons.
coordinate origin has a value of q1=+8.00
mC; the other two have identical
magnitudes, but opposite signs: q2=–5.00
mC and q3=+5.00 mC. (a) Determine the
Now we use Coulomb's Law to find the force: net force (magnitude and direction)
exerted on q1 by the other two charges.
(b) If q1 had a mass of 1.50 g and it were
free to move, what would be its
acceleration?
for some reason i think i did this wrong
abs(F13) = abs(F12)
B.) Since the two charges go from both being neutral to = (8.99*10^9)(8*10^-6)(5*10^-
opposite charges, the force will be repulsive.
6)/(1.3^2)
= 2.13*10^-19
74. A proton and an electron are moving due east in a
constantelectric field that also points due east. (a) Does the F13 will make q1 go up and left, F12
eachexperience an electric force of the same magnitude will make q1 go up and right.
anddirection? (b) What is the direction of the
the left and the right forces cancel,
proton'sacceleration, and what is the direction of the
electron'sacceleration? (c) Is the magnitude of the meaning that the q1 will travel directly
proton'sacceleration greater than, less than, or the same as up??? (by a magnitude of 2*sin(23)*F12
that of theelectron's acceleration? Explain your answer. => 1.66*10^-19)