Answer :
a) [tex]8.93\cdot 10^{-30} N[/tex], downward
b) [tex]1.76\cdot 10^{-17} N[/tex], upward
c) [tex]1.20\cdot 10^{-17} N[/tex], downward
Explanation:
a)
The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.
For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation
[tex]F=mg[/tex]
where
m is the mass of the object
g is the acceleration due to gravity
In this problem, we have:
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electron
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Therefore, the gravitational force on the electron is:
[tex]F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N[/tex]
And the direction is downward, towards the Earth's centre.
b)
The electric force on a charged particle is the force produced by the presence of an electric field.
In particular, this force is:
- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field
- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field
The magnitude of the electric force is given by:
[tex]F=qE[/tex]
where
q is the charge of the particle
E is the strength of the electric field
In this problem:
[tex]q=1.6\cdot 10^{-19} C[/tex] is the charge of the electron
[tex]E=110 N/C[/tex] is the strength of the electric field
Substituting,
[tex]F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N[/tex]
And since the electron has negative charge, the direction of the force is opposite to that of the electric field, so upward.
c)
When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.
The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)
[tex]F=qvB[/tex]
where
q is the charge
v is the velocity of the particle
B is the strength of the magnetic field
In this problem:
[tex]q=1.6\cdot 10^{-19} C[/tex] is the charge of the electron
[tex]v=1.50\cdot 10^6 m/s[/tex] is the velocity
[tex]B=50.0\mu T = 50.0 \cdot 10^{-6} T[/tex] is the strength of the magnetic field
Substituting,
[tex]F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N[/tex]
And the direction can be determined using the right-hand rule:
- Index finger: direction of velocity (eastward)
- Middle finger: direction of magnetic field (northward)
- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward