- #1
ProtoXtype
- 1
- 0
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?
Homework Equations
N = q/e
q = (mgr)/[tex]\Delta[/tex]V
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.