How to Calculate Forces on an Electron at the Equator?

[Pepsi24chevy]

Here reads the problem. the equator, near the surface of the Earth, the magnetic field is approximately 50.0 µT northward, and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron in this environment, assuming the electron has an instantaneous velocity of 7.70 x10^6 m/s directed to the east.


Now for gravitational do i set it up by: (G * m1 * m2) / (d2)? If so what would i use for the distance.

For electric force i assume i set it up by: kq/r^2 ?

And then for magnetic forces i assume i set it up by: qvxB?

Also for the directions, would i use teh right hand rule for each one?

Any suggests towards the right way of doing this would be appreciated. Thanks

 

[LeonhardEuler]

For the gravitational force, if you were to use $$F=\frac{Gm_1m_2}{r^2}$$, then r would be the radius of the earth. However, this is unnecesary since around the surface of the Earth the terms $$\frac{Gm_1}{r^2}$$ are nearly constant and given by $g \approx 9.81m/s^2$, so the force is just mg.

For the electric force, $\vec{F}=q\vec{E}$ would be more appropriate.

Fort the magnetic force you have it right. The right hand rule is the way to find the direction of the magnetic force.

 

[quicknote]

Hello,

Thank you for presenting this problem. I would approach this problem by first identifying the relevant equations and variables, and then setting up the calculations using the given information.

For the gravitational force, you are correct in using the equation F = (G * m1 * m2) / (d^2). In this case, m1 would be the mass of the Earth, m2 would be the mass of the electron (which is approximately 9.11 x 10^-31 kg), and d would be the distance from the center of the Earth to the electron's position (which we can approximate as the radius of the Earth, 6.37 x 10^6 m). Using these values, you can calculate the gravitational force on the electron.

For the electric force, the correct equation is F = k * (q1 * q2) / (r^2), where k is the Coulomb constant (8.99 x 10^9 N*m^2/C^2), q1 is the charge of the Earth (which we can approximate as 0 since it is neutral), q2 is the charge of the electron (which is -1.60 x 10^-19 C), and r is the distance between the electron and the Earth's center (which, again, we can approximate as the radius of the Earth). Using these values, you can calculate the electric force on the electron.

For the magnetic force, the correct equation is F = q * v * B, where q is the charge of the electron, v is its velocity (7.70 x 10^6 m/s directed to the east), and B is the magnetic field strength (50.0 x 10^-6 T). Using these values, you can calculate the magnetic force on the electron.

As for the directions, you are correct in using the right-hand rule for the magnetic force. For the electric force, since the charge of the Earth is neutral, the direction of the force will be zero. And for the gravitational force, since it is an attractive force, the direction will be towards the center of the Earth.

I hope this helps guide you in the right direction for solving this problem. Good luck!