Problem 33 - Physics

Question

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of \(8480 \mathrm{~N} / \mathrm{C}\). The mass of the water drop is \(3.50 \times 10^{-9} \mathrm{~kg} .\) (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?

Step-by-Step Solution

Answer

(a) The charge is positive; the electric field balances gravity upward. (b) Approximately \( 2.53 \times 10^7 \) excess protons.

Step 1: Understanding the Forces

The suspended water drop experiences forces due to gravity and the electric field. Since the drop is motionless, the gravitational force downward must be balanced by the electric force upward. Therefore, the electric force is equal in magnitude and opposite in direction to the gravitational force.

Step 2: Calculating Gravitational Force

The gravitational force on the drop can be calculated using the formula: \( F_g = m \cdot g \), where \( m = 3.50 \times 10^{-9} \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \). Substitute the values: \[ F_g = 3.50 \times 10^{-9} \times 9.81 \approx 3.43 \times 10^{-8} \text{ N}. \]

Step 3: Relate Gravitational Force to Electric Force

Since the forces are balanced, \( F_e = F_g \). The electric force \( F_e = qE \), where \( E = 8480 \text{ N/C} \). Thus, \( qE = F_g \). Substitute \( F_g \) and \( E \) into the equation: \[ q \times 8480 = 3.43 \times 10^{-8}. \] Solve for the charge \( q \).

Step 4: Solving for the Charge

Divide both sides of the equation by \( 8480 \): \[ q = \frac{3.43 \times 10^{-8}}{8480} \approx 4.05 \times 10^{-12} \text{ C}. \] The positive value indicates a positive charge if the electric field direction is opposite to gravity.

Step 5: Determining the Type of Charge

Since the electric field is directed upward and is balancing the gravitational force downward, the charge must be positive to create an upward electric force. Thus, the excess charge is positive.

Step 6: Calculating the Number of Excess Protons or Electrons

Since the charge is positive, it corresponds to excess protons. To find the number of excess protons, use the fundamental charge \( e = 1.60 \times 10^{-19} \text{ C} \): \[ \text{Number of protons} = \frac{4.05 \times 10^{-12}}{1.60 \times 10^{-19}} \approx 2.53 \times 10^7. \] Therefore, about \( 2.53 \times 10^7 \) excess protons reside on the drop.

Key Concepts

Gravitational Force

Gravitational force is the attractive force that pulls objects toward each other. In our scenario, this force pulls the small water drop downward towards the Earth. You can easily calculate the gravitational force if you know the mass of the object and the acceleration due to gravity. The formula is: \[ F_g = m \cdot g \] where:

• \( F_g \) is the gravitational force in newtons (N)
• \( m \) is the mass of the object in kilograms (kg)
• \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \text{ m/s}^2 \) on Earth

So for a water drop with mass \( 3.50 \times 10^{-9} \text{ kg} \), the gravitational force is calculated to be \( 3.43 \times 10^{-8} \text{ N} \). This means that a tiny force is pulling the drop down.

Electric Force

Electric force is what keeps our water drop stationary despite the gravitational pull. This force is generated by an electric field acting on a charged object. Here, since the drop is suspended in air, the upward electric force exactly counteracts the downward gravitational force.The key to understanding electric force lies in its formula: \[ F_e = qE \] where:

• \( F_e \) is the electric force in newtons (N)
• \( q \) is the charge on the object in coulombs (C)
• \( E \) is the electric field strength in newtons per coulomb (N/C)

In our example, the electric field is directed upwards with a strength of \( 8480 \text{ N/C} \). This means if the object is positively charged, it would experience an upward force, which is what happens here. This upward electric force balances the gravitational pull.

Charge Calculation

Calculating the charge on the water drop requires knowing that the electric force balances the gravitational force. Since these forces are equal in magnitude, we set their equations equal:\[ F_g = F_e \]From the formulas \( F_g = m \cdot g \) and \( F_e = qE \), we can establish:\[ q \times 8480 = 3.43 \times 10^{-8} \]To find the charge \( q \), divide both sides by \( 8480 \):\[ q = \frac{3.43 \times 10^{-8}}{8480} \approx 4.05 \times 10^{-12} \text{ C} \]This positive value indicates that the water drop has a positive charge, because the electric field direction is upwards, countering the downward gravitational force.

Excess Protons

The positive charge calculated for the water drop corresponds to an excess of protons. In physics, each proton has a fundamental charge of \( e = 1.60 \times 10^{-19} \text{ C} \). Knowing the total charge \( q \), we can calculate the number of excess protons by dividing the total charge by the charge of a single proton:\[ \text{Number of protons} = \frac{4.05 \times 10^{-12}}{1.60 \times 10^{-19}} \approx 2.53 \times 10^7 \]This means approximately \( 2.53 \times 10^7 \) excess protons reside on the drop, creating the positive charge. Since electrons carry a negative charge, the presence of excess protons results in a net positive charge.