A proton and an electron are moving due east in a constant electric field that also points due east. The electric field has a magnitude of 8.0 ¡Á 104 N/C. Determine the magnitude of the acceleration of the proton and the electron.
Identical point charges of +1.1 ¦ÌC are fixed to three of the four corners of a square. What is the magnitude q of the negative point charge that must be fixed to the fourth corner, so that the charge at the diagonally opposite corner experiences a net force of zero?
F = m a
F = E q same for both where q = electron charge, + for proton, - for electron
m = mass electron for electron
m = mass proton for proton
To determine the magnitude of the acceleration of the proton and the electron, we need to use the equation for the force experienced by a charged particle in an electric field.
The equation is given as:
F = qE
Where:
F = force experienced by the charged particle
q = charge of the particle
E = electric field strength
In this case, both the proton and the electron have the same magnitude of charge (e = 1.6 x 10^-19 C), but opposite signs. The proton has a positive charge, while the electron has a negative charge.
For the proton:
F = qE
F = (1.6 x 10^-19 C) * (8.0 x 10^4 N/C)
F = 1.28 x 10^-14 N
Then, according to Newton's second law, F = ma, where m is the mass of the particle and a is its acceleration. Since the proton's mass is 1.67 x 10^-27 kg:
a = F/m
a = (1.28 x 10^-14 N) / (1.67 x 10^-27 kg)
a ≈ 7.67 x 10^12 m/s^2
Therefore, the magnitude of the acceleration of the proton is approximately 7.67 x 10^12 m/s^2.
For the electron, since it has a negative charge, the direction of the force is opposite to that of the electric field. Therefore, the magnitude of the force experienced by the electron is the same, but the direction is opposite:
F = qE
F = (1.6 x 10^-19 C) * (8.0 x 10^4 N/C)
F = -1.28 x 10^-14 N
Using the mass of the electron (9.11 x 10^-31 kg):
a = F/m
a = (-1.28 x 10^-14 N) / (9.11 x 10^-31 kg)
a ≈ -1.40 x 10^16 m/s^2
Note that the negative sign indicates that the acceleration of the electron is in the opposite direction to that of the electric field.
Therefore, the magnitude of the acceleration of the electron is approximately 1.40 x 10^16 m/s^2.