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Sagot :
Sure, let's go through the problem step-by-step, using the given data systematically to find the radius of a water drop that would remain suspended in an electric field.
### Step 1: Understand the Given Data
- Electron mass, [tex]\( m \)[/tex] = [tex]\( 9.1 \times 10^{-31} \)[/tex] kg
- Electron charge, [tex]\( q \)[/tex] = [tex]\( 1.5 \times 10^{-18} \)[/tex] C
- Electric field strength, [tex]\( E \)[/tex] = 300 V/cm = 30000 V/m (since 1 V/cm = 100 V/m)
- Acceleration due to gravity, [tex]\( g \)[/tex] = 9.81 m/s²
- Density of water, [tex]\( \rho \)[/tex] = 1000 kg/m³
### Step 2: Calculate the Electric Force Acting on the Electron
The force on the electron in the electric field can be computed using the equation:
[tex]\[ F_{\text{electric}} = qE \][/tex]
- Substituting the given values:
[tex]\[ F_{\text{electric}} = 1.5 \times 10^{-18} \, \text{C} \times 30000 \, \text{V/m} \][/tex]
[tex]\[ F_{\text{electric}} = 4.5 \times 10^{-14} \, \text{N} \][/tex]
### Step 3: Calculate the Gravitational Force Acting on the Electron
The gravitational force acting on a particle is given by:
[tex]\[ F_{\text{gravity}} = mg \][/tex]
- Substituting the given values for the mass of the electron and acceleration due to gravity:
[tex]\[ F_{\text{gravity}} = 9.1 \times 10^{-31} \, \text{kg} \times 9.81 \, \text{m/s}^2 \][/tex]
[tex]\[ F_{\text{gravity}} = 8.9271 \times 10^{-30} \, \text{N} \][/tex]
### Step 4: Equating Electric Force and Gravitational Force
To find the equilibrium condition where the drop remains suspended:
[tex]\[ F_{\text{electric}} = F_{\text{gravity}} \][/tex]
Since we are trying to find the radius of the drop, we can rewrite the gravitational force in terms of the water drop's mass and density.
### Step 5: Finding the Mass of Water Drop in Terms of Its Radius
The mass [tex]\( m \)[/tex] of a water drop can be associated with its radius [tex]\( r \)[/tex] considering the density [tex]\( \rho \)[/tex]:
[tex]\[ m = \left(\frac{4}{3} \pi r^3 \right) \rho \][/tex]
### Step 6: Substitute Mass and Find the Radius
We equate the electric force formula and the gravitational force formula to solve for the radius [tex]\( r \)[/tex]:
[tex]\[ qE = \left(\frac{4}{3} \pi r^3 \right) \rho g \][/tex]
Substituting the values:
[tex]\[ 4.5 \times 10^{-14} = \left(\frac{4}{3} \pi r^3 \right) 1000 \times 9.81 \][/tex]
Rearranging to solve for [tex]\( r^3 \)[/tex]:
[tex]\[ r^3 = \frac{4.5 \times 10^{-14}}{(4/3) \pi \times 1000 \times 9.81} \][/tex]
[tex]\[ r^3 = 1.0951028194396469 \times 10^{-18} \, \text{m}^3 \][/tex]
### Step 7: Calculating the Radius
To find the radius [tex]\( r \)[/tex], take the cube root of the result:
[tex]\[ r = \left(1.0951028194396469 \times 10^{-18}\right)^{1/3} \][/tex]
[tex]\[ r = 1.030745938879543 \times 10^{-6} \, \text{m} \][/tex]
### Conclusion
The radius of the water drop that would remain suspended in an electric field of 300 V/cm and is charged with one electron is approximately [tex]\( 1.0307 \times 10^{-6} \)[/tex] meters, or 1.0307 micrometers.
### Step 1: Understand the Given Data
- Electron mass, [tex]\( m \)[/tex] = [tex]\( 9.1 \times 10^{-31} \)[/tex] kg
- Electron charge, [tex]\( q \)[/tex] = [tex]\( 1.5 \times 10^{-18} \)[/tex] C
- Electric field strength, [tex]\( E \)[/tex] = 300 V/cm = 30000 V/m (since 1 V/cm = 100 V/m)
- Acceleration due to gravity, [tex]\( g \)[/tex] = 9.81 m/s²
- Density of water, [tex]\( \rho \)[/tex] = 1000 kg/m³
### Step 2: Calculate the Electric Force Acting on the Electron
The force on the electron in the electric field can be computed using the equation:
[tex]\[ F_{\text{electric}} = qE \][/tex]
- Substituting the given values:
[tex]\[ F_{\text{electric}} = 1.5 \times 10^{-18} \, \text{C} \times 30000 \, \text{V/m} \][/tex]
[tex]\[ F_{\text{electric}} = 4.5 \times 10^{-14} \, \text{N} \][/tex]
### Step 3: Calculate the Gravitational Force Acting on the Electron
The gravitational force acting on a particle is given by:
[tex]\[ F_{\text{gravity}} = mg \][/tex]
- Substituting the given values for the mass of the electron and acceleration due to gravity:
[tex]\[ F_{\text{gravity}} = 9.1 \times 10^{-31} \, \text{kg} \times 9.81 \, \text{m/s}^2 \][/tex]
[tex]\[ F_{\text{gravity}} = 8.9271 \times 10^{-30} \, \text{N} \][/tex]
### Step 4: Equating Electric Force and Gravitational Force
To find the equilibrium condition where the drop remains suspended:
[tex]\[ F_{\text{electric}} = F_{\text{gravity}} \][/tex]
Since we are trying to find the radius of the drop, we can rewrite the gravitational force in terms of the water drop's mass and density.
### Step 5: Finding the Mass of Water Drop in Terms of Its Radius
The mass [tex]\( m \)[/tex] of a water drop can be associated with its radius [tex]\( r \)[/tex] considering the density [tex]\( \rho \)[/tex]:
[tex]\[ m = \left(\frac{4}{3} \pi r^3 \right) \rho \][/tex]
### Step 6: Substitute Mass and Find the Radius
We equate the electric force formula and the gravitational force formula to solve for the radius [tex]\( r \)[/tex]:
[tex]\[ qE = \left(\frac{4}{3} \pi r^3 \right) \rho g \][/tex]
Substituting the values:
[tex]\[ 4.5 \times 10^{-14} = \left(\frac{4}{3} \pi r^3 \right) 1000 \times 9.81 \][/tex]
Rearranging to solve for [tex]\( r^3 \)[/tex]:
[tex]\[ r^3 = \frac{4.5 \times 10^{-14}}{(4/3) \pi \times 1000 \times 9.81} \][/tex]
[tex]\[ r^3 = 1.0951028194396469 \times 10^{-18} \, \text{m}^3 \][/tex]
### Step 7: Calculating the Radius
To find the radius [tex]\( r \)[/tex], take the cube root of the result:
[tex]\[ r = \left(1.0951028194396469 \times 10^{-18}\right)^{1/3} \][/tex]
[tex]\[ r = 1.030745938879543 \times 10^{-6} \, \text{m} \][/tex]
### Conclusion
The radius of the water drop that would remain suspended in an electric field of 300 V/cm and is charged with one electron is approximately [tex]\( 1.0307 \times 10^{-6} \)[/tex] meters, or 1.0307 micrometers.
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