Particles in electric fields

[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.

 

[rwisz]

In part (a), you correctly calculated the number of excess electrons on the sphere by equating the electric force on the sphere to its weight. In part (b), since the potential difference between the plates remains the same, the electric field between the plates also remains the same. This means that the electric force on the sphere is still equal to its weight. However, since the sphere has now lost one excess electron, the electric force on the sphere will decrease by the charge of one electron (1.6x10^-19 C). This will result in a smaller net force on the sphere, leading to a smaller acceleration than before. You can use the equation F=ma to calculate the new acceleration, keeping in mind that the net force on the sphere is now the difference between its weight and the electric force. I hope this helps.