I don't want someone to do this for me-I'm just totally lost and don't even know where to start-I'm totally lost on this one- the earth is surrounded by an electric field with a strenth of 100 N/C at the surface. Its electric properties are the same as a point charge located at its center. How would the gravatational force and electrical force compare foor an electron at earth's surface. Compare magnitude and direction.
I don't know where to even start-this is a high school Physics, not AP, and I have the formulas but I don't know where I find the point charge etc to work with-Any help-thank you
Relax. This is pretty easy. Not even AP
The electrical force on an electron at the Earth's surface is
F_e = E*e ,
where
e is the electron charge and E is the field strength they gave you.
The gravitational force on an electron at the same place is
F_g = m_e*g
where m_e is the mass of the electron and g is the acceleration of gravity.
You should know where to look up e, m_e and g.
Compute and compare the two forces.
Now that doesn't look so hard-
Thanks for breaking it into smaller steps for me
To compare the gravitational force and electrical force on an electron at Earth's surface, you can start by applying the respective formulas for these forces.
1. Gravitational Force:
The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation:
F_grav = (G * m1 * m2) / r^2
Where:
- F_grav is the magnitude of the gravitational force
- G is the gravitational constant (approximately 6.67430 × 10^-11 N (m/kg)^2)
- m1 and m2 are the masses of the two objects
- r is the distance between their centers
In this case, one of the objects is the Earth and the other is the electron. The mass of the Earth is much larger than the mass of the electron, so you can consider the mass of the Earth as the point mass (m1) located at its center, and the electron as the second object (m2) located at the Earth's surface.
2. Electrical Force:
The electrical force between two charged objects can be calculated using Coulomb's Law:
F_elec = (k * |q1 * q2|) / r^2
Where:
- F_elec is the magnitude of the electrical force
- k is Coulomb's constant (approximately 8.9875517923 × 10^9 N(m^2/C^2))
- |q1| and |q2| are the magnitudes of the charges of the two objects
- r is the distance between their centers
In this case, the Earth is treated as a point charge located at its center. So, you need to find the charge (q1) of the Earth. The strength of the electric field at the Earth's surface can provide a clue. Recall that the electric field is defined as the force per unit charge, so E = F_elec / q2. You know that E is 100 N/C, and q2 is the charge of the electron (approximately -1.602 × 10^-19 C).
To find q1, you can rearrange the formula as q1 = (F_elec * r^2) / (k * q2).
Once you have the values for F_grav and F_elec, you can compare their magnitudes and directions.
Remember to use consistent units (such as kilograms, meters, coulombs) in calculations to ensure accurate results.