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A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n?
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A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and r...
Equilibrium of the Nonuniform Disc
To determine the equilibrium position of the nonuniform disc, we need to consider the forces acting on it.
Electric Force
- The disc possesses a charge Q, which creates an electric field around it.
- This electric field exerts a force on the disc due to its charge Q.
Gravitational Force
- The disc has a mass M, which results in a gravitational force acting on it.
- This force acts vertically downwards through the center of mass of the disc.
Normal Force
- Since the disc is placed on a horizontal plane, there is a normal force acting perpendicular to the plane to support the weight of the disc.
Torque Balance
- In order for the disc to be in equilibrium, the sum of the torques acting on it must be zero.
- The torques due to the electric force, gravitational force, and normal force should balance each other out.
Center of Mass
- The nonuniform distribution of mass in the disc will affect the location of its center of mass.
- This will in turn impact the torque calculations and equilibrium position of the disc.
By analyzing the forces and torques acting on the nonuniform disc, we can determine its equilibrium position on the frictionless horizontal plane.
To determine the equilibrium position of the nonuniform disc, we need to consider the forces acting on it.
Electric Force
- The disc possesses a charge Q, which creates an electric field around it.
- This electric field exerts a force on the disc due to its charge Q.
Gravitational Force
- The disc has a mass M, which results in a gravitational force acting on it.
- This force acts vertically downwards through the center of mass of the disc.
Normal Force
- Since the disc is placed on a horizontal plane, there is a normal force acting perpendicular to the plane to support the weight of the disc.
Torque Balance
- In order for the disc to be in equilibrium, the sum of the torques acting on it must be zero.
- The torques due to the electric force, gravitational force, and normal force should balance each other out.
Center of Mass
- The nonuniform distribution of mass in the disc will affect the location of its center of mass.
- This will in turn impact the torque calculations and equilibrium position of the disc.
By analyzing the forces and torques acting on the nonuniform disc, we can determine its equilibrium position on the frictionless horizontal plane.
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A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n?
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A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n?.
A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin non conducting disc of mass M=2kg charge Q=2times10^(-2)C and radius R=(1)/(6)m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate sytem.A nonuniform n?.
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