help with electrostatic charges?

Question

An electron is released a short distance above the surface of the Earth. A second electron directly below it exerts an electrostatic force on the first electron just great enough to cancel the gravitational force on it. How far below the first electron is the second?


Anyone who can help me with this

I used the same formula as GusBsAs
but my answer was 5.0752Km

i took:
m = 9.11e-11Kg
g = 9.81ms-2
e = 1.6e-19Couloumbs
epsilon = 8.85e-12

Answer 1

Since it is near the Earth's surface


Weight = Electric force (apposing the force of gravity)
we write

W=Fe
mg=kq^2/R^2
m- mass of the electron =9.109 E-31 kg
q- charge of an electron= –1.602 E -19 C
k = 8.988 E+9 Nm^2/C^2
g- 9.81 m/s^2
then
R=q sqrt(k/mg)
R= 1.602 E -19 sqrt( 8.988 E+9 /(9.109 E-31 x 9.81))
R=5.08m
It seems you are off by a 1000. No big deal.

Have fun!

Answer 2

since the first electron is released a short distance above the surface of the earth so its distance from the center of the earth would be equal to the earth's radius so :

the gravitational force on the electron = the electrostatic force

G (M1/M2)/ R^2 = K (e^2) / X^2
where G is the gravitational constant , M1 is the mass of the earth M2 is the mass of the electron , R is the radius of the earth ,K is coulomb's constant , e the charge of the electorn , X is the distance between the two electrons
since all the variable are know in the above equation except (X)
which is what you need you can calculate it easily.


you can mail me on:
[e-mail]
for any further questions

Answer 3

e² / (4 π ε0 d²) = m g =>

d = e / √(4 π ε0 m g) = 50.81 µm <= wrong!

Addenda:

Edward is right, the right number is about 5 m.