8 1/2 HRS ON THIS MASTERING PHYSICS WOW
n 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled mg. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 *10^-19 C-the charge of the electron. For this he won the Nobel Prize.
1. If a drop of mass 1.0 * 10^-14 kg remains stationary in an electric field of 2.0 * 10^5 N/C, what is the charge of this drop?
2. If a drop of mass 1.0 * 10^-14kg remains stationary in an electric field of 2.0 * 10^5; N/C}, what is the charge of this drop?
mg = qE
q =mg/E = 1•10^-14•9.8/2•10^5 =
= 4.9•10^-19 C
thanks so much
How many extra electrons are on this particular oil drop (given the presently known charge of the electron)?
1. Well, let's do some calculations and find the charge of this drop. We know that the upward force (qE) is equal to the downward force (mg). Since the drop remains stationary, these forces are balanced.
So, we have qE = mg. Rearranging this equation, we can solve for q:
q = (mg) / E.
Plugging in the given values:
m = 1.0 * 10^-14 kg (the mass of the drop)
g = 9.8 m/s^2 (acceleration due to gravity)
E = 2.0 * 10^5 N/C (electric field strength).
Now, calculate like there's no tomorrow!
q = (1.0 * 10^-14 kg * 9.8 m/s^2) / (2.0 * 10^5 N/C)
Simplifying this expression gives us the charge of the drop. I'll leave the math up to you. Good luck! And remember, you can always drop some oil puns while you calculate to make it more entertaining.
2. Oh wait, I see you accidentally included the same question twice. But don't worry, I've got a great sense of humor and I won't charge you for the mistake! Just let me know if you have any other questions.
To solve this problem, we can use the equation qE = mg, where q is the charge of the drop, E is the electric field strength, m is the mass of the drop, and g is the acceleration due to gravity.
Let's plug in the given values:
m = 1.0 * 10^-14 kg
E = 2.0 * 10^5 N/C
g = 9.8 m/s^2
1. We want to find the charge (q) of the drop. Rearranging the equation, we have q = (mg) / E.
Substituting the values, we get:
q = (1.0 * 10^-14 kg * 9.8 m/s^2) / (2.0 * 10^5 N/C)
= 9.8 * 10^-14 kg m/s^2 / 2.0 * 10^5 N/C
= 4.9 * 10^-19 C
Therefore, the charge of the drop is 4.9 * 10^-19 C.
2. Since the values given in the second part of the question are the same as in the first part, the charge of the drop will also be 4.9 * 10^-19 C.
To answer these questions, we can use the fact that in Millikan's oil-drop experiment, the upward electrical force on a charged oil drop is given by qE, where q is the charge of the drop and E is the electric field strength. This upward electrical force balances the downward force of gravity given by mg, where m is the mass of the drop and g is the acceleration due to gravity.
1. To find the charge of the drop, we can equate the upward electrical force qE to the downward force of gravity mg:
qE = mg
Given:
m = 1.0 * 10^-14 kg
E = 2.0 * 10^5 N/C
g = 9.8 m/s^2 (acceleration due to gravity)
Substituting the given values into the equation, we have:
q * (2.0 * 10^5 N/C) = (1.0 * 10^-14 kg) * (9.8 m/s^2)
Now we can solve for q:
q = [(1.0 * 10^-14 kg) * (9.8 m/s^2)] / (2.0 * 10^5 N/C)
= 4.9 * 10^-20 C
Therefore, the charge of the drop is 4.9 * 10^-20 C.
2. In this question, it seems that there is a mistake in the formatting, but assuming the correct information is as follows:
m = 1.0 * 10^-14 kg
E = 2.0 * 10^5 N/C
We can follow the same steps as in question 1 to find the charge of the drop:
q * (2.0 * 10^5 N/C) = (1.0 * 10^-14 kg) * (9.8 m/s^2)
Solving for q:
q = [(1.0 * 10^-14 kg) * (9.8 m/s^2)] / (2.0 * 10^5 N/C)
= 4.9 * 10^-20 C
Therefore, the charge of the drop is 4.9 * 10^-20 C.
It seems that the same values were given in both questions, so the result for both questions is the same.