How Many Extra Electrons Does a Latex Sphere Have in an Electric Field?

[ProtoXtype]

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)/\DeltaV

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
\DeltaV = 3.00 x 10^2 V

q = (mgr)/\DeltaV
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:

\DeltaEe = \DeltaEk
q\DeltaV = mv^2/2
q = mv^2/2\DeltaV
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.

 

[alphysicist]

Hi ProtoXtype,

Your approach in attempt 1 looks okay to me; however I believe you have a few errors. When you solved for q you say you got 1.421 x 10^-16; this exponent -16 here does not look right; I think you just misread your calculator.

A couple of lines below that you wrote the charge on an electron as 1.6 x 10^-16, but it looks like maybe you used the correct value in your calculation.


For your attempt 2, the first equation is not correct as it leaves out the change in gravitational potential energy. If you put that in and also set the change in kinetic energy = 0 (since the object is moving at constant speed) you'll find yourself with the expression you used in attempt 1.