In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is positive. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0 × 10-11 kg gets suspended between the plates. (e = 1.6 × 10-19 C)
(a) What is the charge on the oil drop?
(b) Is there an excess or deficit of electrons on the oil drop? How many electrons are in excess or deficit?
(a) The charge on the oil drop is 1.5 × 10-19 C.
(b) There is an excess of electrons on the oil drop. There are 9.375 × 1010 electrons in excess.
To determine the charge on the oil drop in the Millikan's oil drop experiment, we can use the following formula:
Charge (q) = mass (m) * acceleration due to gravity (g) / electric field (E)
(a) Calculate the charge on the oil drop:
Given values:
Mass (m) = 2.0 × 10^-11 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Electric field (E) = V/d = 240.0 V / 0.0200 m = 12000 V/m
Substituting the values into the formula:
Charge (q) = (2.0 × 10^-11 kg) * (9.8 m/s^2) / (12000 V/m)
= 1.63 × 10^-19 C
Therefore, the charge on the oil drop is approximately 1.63 × 10^-19 C.
(b) To determine if there is an excess or deficit of electrons on the oil drop, we can compare the charge on the oil drop with the elementary charge of an electron (e).
Since the charge on the oil drop is 1.63 × 10^-19 C and the elementary charge (e) is 1.6 × 10^-19 C, we can conclude that the oil drop has an excess of electrons.
To calculate the number of excess electrons, we can use the formula:
Number of excess or deficit electrons = charge on the oil drop / elementary charge (e)
Number of excess or deficit electrons = (1.63 × 10^-19 C) / (1.6 × 10^-19 C)
= 1.02
Therefore, the oil drop has approximately 1 electron in excess.
To determine the charge on the oil drop in Millikan's oil drop experiment, we can use the formula:
q = mg + ne
Where:
q is the charge on the oil drop
m is the mass of the oil drop
g is the acceleration due to gravity
n is the number of excess or deficit electrons
e is the charge of a single electron
(a) To find the charge on the oil drop, we need to calculate the number of excess or deficit electrons first.
1. Calculate the weight of the oil drop:
Weight (mg) = mass (m) × gravitational acceleration (g)
Given: mass (m) = 2.0 × 10^(-11) kg
Gravitational acceleration (g) = 9.8 m/s^2
Weight (mg) = (2.0 × 10^(-11) kg) × (9.8 m/s^2)
2. Calculate the number of excess or deficit electrons:
n = (q - mg) / e
Here, q is unknown, so we'll substitute "q" with "mg + ne" from the initial formula.
n = ((mg + ne) - mg) / e
Note: Since the oil drop is suspended, the net force on it is zero, thus weight (mg) is balanced by the electrostatic force (qE) between the plates.
n = ne / e, simplifying the equation
Given: e = 1.6 × 10^(-19) C
Let's plug in the values:
n = ((2.0 × 10^(-11) kg) × (9.8 m/s^2)) / (1.6 × 10^(-19) C)
Now, simplify and calculate the value of n.
Once we have the value of n, we can calculate the charge on the oil drop (q) using the formula:
q = mg + ne
(b) To determine if there is an excess or deficit of electrons, we need to evaluate the sign of "q". If "q" is positive, it implies an excess of electrons, and if "q" is negative, it implies a deficit of electrons. The absolute value of "q" will give us the number of excess or deficit electrons.