Calculating Charge on Drop with Excess Electrons

[Masafi]

Oil drop weighs 5.8 x 10^(-15)N. It is suspended in a uniform field produced by a pf of 363V between 2 parallel plates 6cm apart.

1. Calculate charge on drop.

2. How many excess electrons does it carry?

My question is, why are the electrons excess? The answer is 6, which is basically the answer to 1 divided by the charge of an electron. Why are all the elctrons excess?

 

[Stonebridge]

In a neutral drop there would be an equal number of positive (proton) and negative (electron) charges.
For the drop to be charged, there must be more - than +. Hence the term "excess".

 

[kerimek]

I would like to clarify that the term "excess electrons" refers to the fact that the oil drop has a net negative charge, meaning it has more electrons than protons. This can occur due to various factors such as friction, rubbing, or transfer of electrons from another source.

To calculate the charge on the drop, we can use the formula Q = Fd, where Q is the charge, F is the force (weight) acting on the drop, and d is the distance between the plates. Plugging in the given values, we get Q = (5.8 x 10^(-15)N)(0.06m) = 3.48 x 10^(-16)C.

To determine the number of excess electrons, we need to know the charge of one electron, which is 1.6 x 10^(-19)C. Dividing the total charge by the charge of one electron, we get 3.48 x 10^(-16)C / (1.6 x 10^(-19)C) = 2.175 x 10^3 excess electrons.

It is important to note that the number of excess electrons may not always be a whole number, as it depends on the exact charge and size of the oil drop. Additionally, the number of excess electrons can change if the conditions (such as the electric field or distance between plates) are altered.

I hope this answer addresses your question and clarifies the concept of excess electrons. As scientists, it is crucial to be precise and accurate in our use of terminology.