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**1. Homework Statement**

A 2.9 x 10^-3 kg latex sphere is moving up, at a constant velocity of 0.013 mm/s, between two large plates. The plates are 1.500 cm apart, with a potential difference of 3.00 x 10^2 V between them. How many extra electrons does the sphere have?

**2. Homework Equations**

N = q/e

q = (mgr)/[tex]\Delta[/tex]V

**3. The Attempt at a Solution**

Attempt 1:

m = 2.9 x 10^-3 kg

v = 0.013 mm/s = 1.3 x 10^-5 m/s

r = 1.500 cm = 0.015 m

[tex]\Delta[/tex]V = 3.00 x 10^2 V

q = (mgr)/[tex]\Delta[/tex]V

q = [ (2.9 x 10^-3 kg)(9.80 m/s^2)(0.015 m) ]/ 3.00 x 10^2 V

q = 1.421 x 10^-16 C

N = q/e

N = (1.421 x 10^-16 C )/(1.602 x 10^-16 C)

N = 8.9 x 10^2 extra electrons

Attempt 2:

[tex]\Delta[/tex]Ee = [tex]\Delta[/tex]Ek

q[tex]\Delta[/tex]V = mv^2/2

q = mv^2/2[tex]\Delta[/tex]V

q = [ (2.9 x 10^-3 kg)(1.3 x 10^-5 m/s)^2 ]/ [ 2(3.00 x 10^2 V) ]

q = 8.168 x 10^-26 C

N = q/e

N = (8.168 x 10^-26 C )/(1.602 x 10^-16 C)

N = 5.1 x 10^-7 extra electrons

Not too sure if what I have done so far is right or not. But any help will be appreciated.