How Many Excess Electrons Does a Charged Sphere Have?

[ishterz]

Homework Statement

An negatively charged sphere of mass 3.3x10^-15 kg is held at rest b/w two parallel plates separated by 5.0mm when the potential difference between them is 170V. How many excess electrons are on the sphere?

B) The charged sphere suddenly loses one of its excess electrons. The p.d b/w the plates remains the same. Describe the initial motion of the sphere

Homework Equations

qv/d=mg

The Attempt at a Solution

I got part (a) as 6 excess electrons. I equated the electric field to the weight found the charge and divided it by 1.6X10^-19.

However, I am having trouble with part (b) ; the answer given also gives the value initial acceleration. I tried finding the accerlation by F=ma but didn't get the answer.

Please help
Thanks :)

 

[Redbelly98]

I agree with the 6 electrons answer for part (a).

... I am having trouble with part (b) ; the answer given also gives the value initial acceleration. I tried finding the accerlation by F=ma but didn't get the answer.

Your approach looks correct. Can you show in more detail what you did?

 

[ishterz]

I tried finding the force due to the electric field:
170/.005 x 6 x 1.6X10^-19 (I did 6 because there were 6 excess electrons so that means there are 7 in total)

=3.264X10-14

Then I tried subtracting the weight of 6 electrons from it.
Weight = 9.1x10^-31 x 6 x 9.81

Resultant force = 3.260X10-14

Acceleration = 3.260X10-14/ (9.1 x 10^-31 X 6)
= 5.9 X 10^15

I doubt my method is right since the right answer is "the charged particle moves downwards with inital acceleration 1.6 ms^-2 "

I reckon maybe it has something to do with equations of uniform acceleration?

 

[Redbelly98]

I'll try to clear up some problems with your method.

I tried finding the force due to the electric field:
170/.005 x 6 x 1.6X10^-19 (I did 6 because there were 6 excess electrons so that means there are 7 in total)

=3.264X10-14

Well, I agree that is the force due to the electric field when there are 6 excess electrons. But how many excess electrons are left when, as the problem statement says, "the charged sphere suddenly loses one of its excess electrons"?

Then I tried subtracting the weight of 6 electrons from it.
Weight = 9.1x10^-31 x 6 x 9.81

Resultant force = 3.260X10-14

The gravitational force pulls down on the sphere's entire 3.3x10^-15 kg mass, not just the excess electrons.

Acceleration = 3.260X10-14/ (9.1 x 10^-31 X 6)
= 5.9 X 10^15

I doubt my method is right since the right answer is "the charged particle moves downwards with inital acceleration 1.6 ms^-2 "

I reckon maybe it has something to do with equations of uniform acceleration?

Judging from the answer given, it seems to be more of a Newton's 2nd Law, F=ma, type of problem. You just need to account for the two forces (electric and gravitational) properly, in order to find the net force F.