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To solve the problem of balancing the force of gravity on an oil drop with an electric field, let’s break it down step by step:
1. Convert the mass from milligrams to kilograms:
Given mass [tex]\( \text{mass} = 50 \, \text{mg} \)[/tex]. To convert milligrams to kilograms:
[tex]\[ \text{mass} = 50 \times 10^{-6} \, \text{kg} = 0.00005 \, \text{kg} \][/tex]
2. Convert the charge from microcoulombs to coulombs:
Given charge [tex]\( \text{charge} = -5 \, \mu \text{C} \)[/tex]. To convert microcoulombs to coulombs:
[tex]\[ \text{charge} = -5 \times 10^{-6} \, \text{C} = -0.000005 \, \text{C} \][/tex]
3. Calculate the force of gravity acting on the oil drop:
The force of gravity can be calculated using:
[tex]\[ F_{\text{gravity}} = m \cdot g \][/tex]
where [tex]\( m \)[/tex] is the mass of the oil drop and [tex]\( g \)[/tex] is the acceleration due to gravity (given as [tex]\( 9.8 \, \text{m/s}^{2} \)[/tex]).
[tex]\[ F_{\text{gravity}} = 0.00005 \, \text{kg} \times 9.8 \, \text{m/s}^{2} = 0.00049 \, \text{N} \][/tex]
4. Calculate the electric field required to balance this gravitational force:
To balance the force of gravity, the electric force [tex]\( F_{\text{electric}} \)[/tex] must be equal in magnitude but opposite in direction to the gravitational force. The electric field [tex]\( E \)[/tex] is given by:
[tex]\[ F_{\text{electric}} = q \cdot E \][/tex]
Since [tex]\( F_{\text{electric}} \)[/tex] must equal [tex]\( F_{\text{gravity}} \)[/tex] in magnitude:
[tex]\[ q \cdot E = F_{\text{gravity}} \][/tex]
Solving for [tex]\( E \)[/tex]:
[tex]\[ E = \frac{F_{\text{gravity}}}{|q|} \][/tex]
Substituting in the values:
[tex]\[ E = \frac{0.00049 \, \text{N}}{0.000005 \, \text{C}} = 98 \, \text{N/C} \][/tex]
5. Determine the direction of the electric field:
The oil drop has a negative charge ([tex]\( -5 \mu \text{C} \)[/tex]). To counteract the downward force of gravity with an electric field, the force exerted by the electric field on the negatively charged oil drop must be upward. Hence, the electric field itself must be directed upwards.
Thus, the strength of the electric field required to balance the oil drop is [tex]\( \boxed{98 \, \text{N/C} \, \text{upwards}} \)[/tex]. Therefore, the correct answer is:
[tex]\[ a. \, \boxed{98 \, \text{N/C} \, \text{upwards}} \][/tex]
1. Convert the mass from milligrams to kilograms:
Given mass [tex]\( \text{mass} = 50 \, \text{mg} \)[/tex]. To convert milligrams to kilograms:
[tex]\[ \text{mass} = 50 \times 10^{-6} \, \text{kg} = 0.00005 \, \text{kg} \][/tex]
2. Convert the charge from microcoulombs to coulombs:
Given charge [tex]\( \text{charge} = -5 \, \mu \text{C} \)[/tex]. To convert microcoulombs to coulombs:
[tex]\[ \text{charge} = -5 \times 10^{-6} \, \text{C} = -0.000005 \, \text{C} \][/tex]
3. Calculate the force of gravity acting on the oil drop:
The force of gravity can be calculated using:
[tex]\[ F_{\text{gravity}} = m \cdot g \][/tex]
where [tex]\( m \)[/tex] is the mass of the oil drop and [tex]\( g \)[/tex] is the acceleration due to gravity (given as [tex]\( 9.8 \, \text{m/s}^{2} \)[/tex]).
[tex]\[ F_{\text{gravity}} = 0.00005 \, \text{kg} \times 9.8 \, \text{m/s}^{2} = 0.00049 \, \text{N} \][/tex]
4. Calculate the electric field required to balance this gravitational force:
To balance the force of gravity, the electric force [tex]\( F_{\text{electric}} \)[/tex] must be equal in magnitude but opposite in direction to the gravitational force. The electric field [tex]\( E \)[/tex] is given by:
[tex]\[ F_{\text{electric}} = q \cdot E \][/tex]
Since [tex]\( F_{\text{electric}} \)[/tex] must equal [tex]\( F_{\text{gravity}} \)[/tex] in magnitude:
[tex]\[ q \cdot E = F_{\text{gravity}} \][/tex]
Solving for [tex]\( E \)[/tex]:
[tex]\[ E = \frac{F_{\text{gravity}}}{|q|} \][/tex]
Substituting in the values:
[tex]\[ E = \frac{0.00049 \, \text{N}}{0.000005 \, \text{C}} = 98 \, \text{N/C} \][/tex]
5. Determine the direction of the electric field:
The oil drop has a negative charge ([tex]\( -5 \mu \text{C} \)[/tex]). To counteract the downward force of gravity with an electric field, the force exerted by the electric field on the negatively charged oil drop must be upward. Hence, the electric field itself must be directed upwards.
Thus, the strength of the electric field required to balance the oil drop is [tex]\( \boxed{98 \, \text{N/C} \, \text{upwards}} \)[/tex]. Therefore, the correct answer is:
[tex]\[ a. \, \boxed{98 \, \text{N/C} \, \text{upwards}} \][/tex]
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