A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 12300 N/C. The mass of the water drop is 4.71 × 10-9 kg. How many excess electrons or protons reside on the drop?

To determine the number of excess electrons or protons on the water drop, we need to calculate the charge on the drop first.

Step 1: Find the electric force on the water drop.
The electric force can be calculated using the equation: F = qE, where F is the electric force, q is the charge, and E is the electric field.
Given: E = 12300 N/C
We need to find q.

Step 2: Calculate the electric force on the water drop.
The electric force can be calculated using the equation: F = ma, where F is the force, m is the mass, and a is the acceleration.
Given: m = 4.71 × 10^-9 kg
We'll use the formula F = ma to calculate the electric force.

F = ma
F = (4.71 × 10^-9 kg)(9.8 m/s^2)
F ≈ 4.6158 × 10^-8 N

Step 3: Calculate the charge on the water drop.
Using the equation F = qE, we can rearrange it to solve for q:
q = F / E

q = (4.6158 × 10^-8 N) / (12300 N/C)
q ≈ 3.7524 × 10^-12 C

Step 4: Calculate the number of excess electrons or protons.
The elementary charge of an electron is approximately 1.6 × 10^-19 C.
To find the number of excess electrons (n), we divide the charge (q) by the elementary charge:
n = q / (1.6 × 10^-19 C)

n ≈ (3.7524 × 10^-12 C) / (1.6 × 10^-19 C)
n ≈ 2.34515 × 10^7

Therefore, approximately 23,451,500 excess electrons or protons reside on the water drop.

To determine the number of excess electrons or protons residing on the water drop, we can use the concept of electric force and the charge of a single electron or proton.

First, let's calculate the electric force acting on the water drop. The electric force can be calculated using the equation:

Electric Force (F) = Electric Field (E) * Charge (Q)

Given:
Electric Field (E) = 12300 N/C
Mass of the water drop (m) = 4.71 × 10^-9 kg

To find the charge (Q), we need to know the magnitude of the electric charge on a single electron or proton. The electron has a charge of -1.6 × 10^-19 C, while the proton has a charge of +1.6 × 10^-19 C. Either can be chosen, as they have equal magnitudes but opposite signs.

Let's calculate the electric force acting on the water drop using the given electric field:

F = E * Q

Since the water drop is suspended motionless in the electric field, the electric force acting on it is equal to the gravitational force:

F = m * g

Where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Equating the two forces:

E * Q = m * g

Now, rearrange the equation to solve for the charge (Q):

Q = (m * g) / E

Substitute the given values into the equation:

Q = (4.71 × 10^-9 kg * 9.8 m/s^2) / 12300 N/C

Note: We can use either the magnitude of the electron charge or the proton charge, as mentioned earlier.

By using the electron charge (-1.6 × 10^-19 C), the calculation becomes:

Q = (4.71 × 10^-9 kg * 9.8 m/s^2) / 12300 N/C
≈ -3.764 × 10^-19 C

Since the charge of one electron is -1.6 × 10^-19 C, we can find the number of electrons on the drop by dividing the total charge by the charge of a single electron:

Number of electrons = Q / (-1.6 × 10^-19 C)
≈ (-3.764 × 10^-19 C) / (-1.6 × 10^-19 C)
≈ 2.352

Rounding the result to the nearest whole number, we can conclude that approximately 2 excess electrons reside on the water drop.