Question

At the Equator near Earth’s surface, the magnetic field is approximately 82.2 µT northward and the electric field is about 143 N/C downward in fair weather. A electron travels with an instantaneous velocity of 2.42 × 106 m/s directed to the east in this environment. The acceleration of gravity is 9.8 m/s 2 . Find the magnitude of the gravitational force on the electron. Answer in units of N.

Answer 1

Answer:

Magnitude of gravitational force of the electron= 3.74×10^12N

Explanation:

Felectron= Fgravitational

Therefore:

Felectron/Fgravitational = kq^2/r^2 ×(r^2/Gm^2)

= kq^2/Gm^2

Where G= gravitational constant

m = mass of electron=9.1×10^-31

K= 8.9x10^9

q= 1.6×10^-19

Magnitude of gravitational force= (8.9×10^9)×(1.6×10^-19)^2/(6.7×10^-11)×(9.1×10^-31)

=( 2.2784×10^-28) / ( 6.097×10^-41)

= 3.74×10^12N