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An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy?
Most Upvoted Answer
An oil drop carrying six electronic charges and having a mass of 1.6 ×...
1 st case ;
there is a only a gravitational force to balance drag force
F = mg
2 nd case ;
drop is moving upward so drag force will be in downward direction
F = mg = 6eE from the 1 st case we have,
F = mg
2mg = 6eE
E = mg/3 e = 1.6 × 10-15×10/3×1.6×10-19
= 3.3×10 4NC-1
there is a only a gravitational force to balance drag force
F = mg
2 nd case ;
drop is moving upward so drag force will be in downward direction
F = mg = 6eE from the 1 st case we have,
F = mg
2mg = 6eE
E = mg/3 e = 1.6 × 10-15×10/3×1.6×10-19
= 3.3×10 4NC-1
Community Answer
An oil drop carrying six electronic charges and having a mass of 1.6 ×...
Analysis:
The force acting on the oil drop in the downward direction is the gravitational force (mg) and the force due to air resistance. When the drop reaches terminal velocity, the force due to air resistance is equal in magnitude to the gravitational force.
Force Balance:
- When the drop is moving downward with terminal velocity, the net force on the drop is zero.
- The force due to air resistance is given by the Stokes' law: F = 6πηrv, where η is the viscosity of the medium, r is the radius of the drop, and v is the velocity of the drop.
- The gravitational force is given by F = mg, where m is the mass of the drop and g is the acceleration due to gravity.
- At terminal velocity, F = 6πηrv = mg.
Electric Field Requirement:
- When the drop is moving upward with the same speed as it was moving downward, an additional force due to the electric field is required to counteract the gravitational force.
- The electric force acting on the drop is given by F = qE, where q is the charge on the drop and E is the electric field.
- To make the drop move upward with the same speed, the electric force should be equal in magnitude to the gravitational force: qE = mg.
- Substituting the values of q and m, we get E = mg/q.
Therefore, the magnitude of the vertical electric field required to make the drop move upward with the same speed as it was moving downward is E = (mg)/q.
The force acting on the oil drop in the downward direction is the gravitational force (mg) and the force due to air resistance. When the drop reaches terminal velocity, the force due to air resistance is equal in magnitude to the gravitational force.
Force Balance:
- When the drop is moving downward with terminal velocity, the net force on the drop is zero.
- The force due to air resistance is given by the Stokes' law: F = 6πηrv, where η is the viscosity of the medium, r is the radius of the drop, and v is the velocity of the drop.
- The gravitational force is given by F = mg, where m is the mass of the drop and g is the acceleration due to gravity.
- At terminal velocity, F = 6πηrv = mg.
Electric Field Requirement:
- When the drop is moving upward with the same speed as it was moving downward, an additional force due to the electric field is required to counteract the gravitational force.
- The electric force acting on the drop is given by F = qE, where q is the charge on the drop and E is the electric field.
- To make the drop move upward with the same speed, the electric force should be equal in magnitude to the gravitational force: qE = mg.
- Substituting the values of q and m, we get E = mg/q.
Therefore, the magnitude of the vertical electric field required to make the drop move upward with the same speed as it was moving downward is E = (mg)/q.
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An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy?
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An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy?.
An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An oil drop carrying six electronic charges and having a mass of 1.6 × 10^-12 g. falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy?.
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