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Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]
- a)12 cm
- b)10 cm
- c)8 cm
- d)5 cm
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Two identical charged particles each having a mass 10 g and charge 2.0...
According to given information:
Putting the values, we get
L = 12 cm
L = 12 cm
Most Upvoted Answer
Two identical charged particles each having a mass 10 g and charge 2.0...
Given data:
- Mass of each particle, m = 10 g = 0.01 kg
- Charge of each particle, q = 2.0 x 10^-7 C
- Coefficient of friction, μ = 0.25
- Acceleration due to gravity, g = 10 m/s^2
Equilibrium Condition:
The particles are in limited equilibrium when the electrostatic force between them is balanced by the frictional force.
Electrostatic Force:
The electrostatic force between the particles can be calculated using Coulomb's law:
F = k * (q^2) / r^2
where k is the Coulomb constant, q is the charge of each particle, and r is the separation between the particles.
Frictional Force:
The frictional force acting on each particle can be calculated as:
F_friction = μ * m * g
Equating the Forces:
At equilibrium, the electrostatic force is equal to the frictional force:
k * (q^2) / L^2 = μ * m * g
Finding the Separation L:
Substitute the given values and solve for L:
L = sqrt((k * (q^2)) / (μ * m * g))
Calculating the value of L:
Given:
k = 9 x 10^9 Nm^2/C^2
Substitute the values and calculate the separation:
L = sqrt((9 x 10^9 * (2.0 x 10^-7)^2) / (0.25 * 0.01 * 10))
L = sqrt((36 x 10^-9) / (0.25 x 0.1))
L = sqrt(0.144)
L = 0.12 m = 12 cm
Therefore, the value of L is 12 cm. Hence, option (a) is correct.
- Mass of each particle, m = 10 g = 0.01 kg
- Charge of each particle, q = 2.0 x 10^-7 C
- Coefficient of friction, μ = 0.25
- Acceleration due to gravity, g = 10 m/s^2
Equilibrium Condition:
The particles are in limited equilibrium when the electrostatic force between them is balanced by the frictional force.
Electrostatic Force:
The electrostatic force between the particles can be calculated using Coulomb's law:
F = k * (q^2) / r^2
where k is the Coulomb constant, q is the charge of each particle, and r is the separation between the particles.
Frictional Force:
The frictional force acting on each particle can be calculated as:
F_friction = μ * m * g
Equating the Forces:
At equilibrium, the electrostatic force is equal to the frictional force:
k * (q^2) / L^2 = μ * m * g
Finding the Separation L:
Substitute the given values and solve for L:
L = sqrt((k * (q^2)) / (μ * m * g))
Calculating the value of L:
Given:
k = 9 x 10^9 Nm^2/C^2
Substitute the values and calculate the separation:
L = sqrt((9 x 10^9 * (2.0 x 10^-7)^2) / (0.25 * 0.01 * 10))
L = sqrt((36 x 10^-9) / (0.25 x 0.1))
L = sqrt(0.144)
L = 0.12 m = 12 cm
Therefore, the value of L is 12 cm. Hence, option (a) is correct.
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Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]a)12 cmb)10 cmc)8 cmd)5 cmCorrect answer is option 'A'. Can you explain this answer?
Question Description
Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]a)12 cmb)10 cmc)8 cmd)5 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]a)12 cmb)10 cmc)8 cmd)5 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]a)12 cmb)10 cmc)8 cmd)5 cmCorrect answer is option 'A'. Can you explain this answer?.
Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]a)12 cmb)10 cmc)8 cmd)5 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]a)12 cmb)10 cmc)8 cmd)5 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical charged particles each having a mass 10 g and charge 2.0 x 10-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms-2]a)12 cmb)10 cmc)8 cmd)5 cmCorrect answer is option 'A'. Can you explain this answer?.
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