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.

Question

Physics

Mike Velez

27 September, 16:23

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.

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Mccormick · Answer 1

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