Answer :
Sure, let's work through the problem step-by-step.
Given Data:
- Charge of one electron, [tex]\( e = 1.6 \times 10^{-19} \, \text{C} \)[/tex]
- Distance between the plates, [tex]\( d = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)[/tex]
- Mass of the oil drop, [tex]\( m = 1.31 \times 10^{-14} \, \text{kg} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
- Number of electrons attached to the oil drop, [tex]\( n = 2 \)[/tex]
To Find:
The potential difference [tex]\( V \)[/tex] in volts required to keep the oil drop in equilibrium.
Step-by-Step Solution:
1. Calculate the total charge on the oil drop:
[tex]\[ \text{Total charge} = n \times e \][/tex]
Substituting the given values:
[tex]\[ \text{Total charge} = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \][/tex]
2. Calculate the weight of the oil drop:
[tex]\[ \text{Weight} (W) = m \times g \][/tex]
Substituting the given values:
[tex]\[ \text{Weight} = 1.31 \times 10^{-14} \, \text{kg} \times 9.81 \, \text{m/s}^2 = 1.28511 \times 10^{-13} \, \text{N} \][/tex]
3. Relate the weight to the electric force:
In equilibrium, the electric force ([tex]\( F_{\text{electric}} \)[/tex]) will balance the weight of the oil drop:
[tex]\[ F_{\text{electric}} = W \][/tex]
The electric force can also be expressed as:
[tex]\[ F_{\text{electric}} = \frac{V \times \text{Total charge}}{d} \][/tex]
4. Create the equation for the potential difference:
Equating the forces:
[tex]\[ W = \frac{V \times \text{Total charge}}{d} \][/tex]
Solving for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{W \times d}{\text{Total charge}} \][/tex]
Substituting the known values:
[tex]\[ V = \frac{1.28511 \times 10^{-13} \, \text{N} \times 5 \times 10^{-3} \, \text{m}}{3.2 \times 10^{-19} \, \text{C}} \][/tex]
5. Calculate the potential difference:
[tex]\[ V = \frac{6.42555 \times 10^{-16}}{3.2 \times 10^{-19}} \][/tex]
Simplifying the fraction:
[tex]\[ V \approx 2007.98 \, \text{V} \][/tex]
So, the potential difference necessary to be maintained between the two horizontal conducting plates is approximately [tex]\( 2007.98 \, \text{V} \)[/tex].
Given Data:
- Charge of one electron, [tex]\( e = 1.6 \times 10^{-19} \, \text{C} \)[/tex]
- Distance between the plates, [tex]\( d = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)[/tex]
- Mass of the oil drop, [tex]\( m = 1.31 \times 10^{-14} \, \text{kg} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
- Number of electrons attached to the oil drop, [tex]\( n = 2 \)[/tex]
To Find:
The potential difference [tex]\( V \)[/tex] in volts required to keep the oil drop in equilibrium.
Step-by-Step Solution:
1. Calculate the total charge on the oil drop:
[tex]\[ \text{Total charge} = n \times e \][/tex]
Substituting the given values:
[tex]\[ \text{Total charge} = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \][/tex]
2. Calculate the weight of the oil drop:
[tex]\[ \text{Weight} (W) = m \times g \][/tex]
Substituting the given values:
[tex]\[ \text{Weight} = 1.31 \times 10^{-14} \, \text{kg} \times 9.81 \, \text{m/s}^2 = 1.28511 \times 10^{-13} \, \text{N} \][/tex]
3. Relate the weight to the electric force:
In equilibrium, the electric force ([tex]\( F_{\text{electric}} \)[/tex]) will balance the weight of the oil drop:
[tex]\[ F_{\text{electric}} = W \][/tex]
The electric force can also be expressed as:
[tex]\[ F_{\text{electric}} = \frac{V \times \text{Total charge}}{d} \][/tex]
4. Create the equation for the potential difference:
Equating the forces:
[tex]\[ W = \frac{V \times \text{Total charge}}{d} \][/tex]
Solving for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{W \times d}{\text{Total charge}} \][/tex]
Substituting the known values:
[tex]\[ V = \frac{1.28511 \times 10^{-13} \, \text{N} \times 5 \times 10^{-3} \, \text{m}}{3.2 \times 10^{-19} \, \text{C}} \][/tex]
5. Calculate the potential difference:
[tex]\[ V = \frac{6.42555 \times 10^{-16}}{3.2 \times 10^{-19}} \][/tex]
Simplifying the fraction:
[tex]\[ V \approx 2007.98 \, \text{V} \][/tex]
So, the potential difference necessary to be maintained between the two horizontal conducting plates is approximately [tex]\( 2007.98 \, \text{V} \)[/tex].
