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The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy?
Most Upvoted Answer
The force acting between two point charges kept at a certain distance ...
Its b 4phy
F=kQ2/r2=phy/4
r2=kQ2×4/phy
when Q is dounled and distance is halved
f=k4×Q2 ×phy×4/kQ2×4
f=4phy
F=kQ2/r2=phy/4
r2=kQ2×4/phy
when Q is dounled and distance is halved
f=k4×Q2 ×phy×4/kQ2×4
f=4phy
Community Answer
The force acting between two point charges kept at a certain distance ...
Given:
- Force between two point charges at a certain distance = phy/4
To find:
- Force between the same charges when charge magnitudes are doubled and distance is halved.
Solution:
Step 1: Understanding the given scenario
- Let's assume the two point charges as q1 and q2.
- The force between the charges is given by Coulomb's law: F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
- Given that the force is phy/4, we can write the equation as phy/4 = k * (q1 * q2) / r^2.
Step 2: Doubling the charge magnitudes and halving the distance
- Now, let's double the magnitudes of the charges: q1' = 2q1 and q2' = 2q2.
- Also, let's halve the distance between the charges: r' = r/2.
Step 3: Finding the new force
- Using Coulomb's law with the new magnitudes and distance, the force can be calculated as: F' = k * (q1' * q2') / r'^2.
- Substituting the values: F' = k * (2q1 * 2q2) / (r/2)^2 = (4 * k * q1 * q2) / (r^2/4) = 16 * (k * q1 * q2) / r^2.
Step 4: Comparing the forces
- Comparing the new force F' with the given force phy/4, we have: F' = 16 * (k * q1 * q2) / r^2 = 16 * (phy/4) = 4phy.
Conclusion:
- The force acting between the charges, when the magnitudes are doubled and the distance is halved, is 4phy.
- Therefore, the correct option is (b) 4phy.
- Force between two point charges at a certain distance = phy/4
To find:
- Force between the same charges when charge magnitudes are doubled and distance is halved.
Solution:
Step 1: Understanding the given scenario
- Let's assume the two point charges as q1 and q2.
- The force between the charges is given by Coulomb's law: F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
- Given that the force is phy/4, we can write the equation as phy/4 = k * (q1 * q2) / r^2.
Step 2: Doubling the charge magnitudes and halving the distance
- Now, let's double the magnitudes of the charges: q1' = 2q1 and q2' = 2q2.
- Also, let's halve the distance between the charges: r' = r/2.
Step 3: Finding the new force
- Using Coulomb's law with the new magnitudes and distance, the force can be calculated as: F' = k * (q1' * q2') / r'^2.
- Substituting the values: F' = k * (2q1 * 2q2) / (r/2)^2 = (4 * k * q1 * q2) / (r^2/4) = 16 * (k * q1 * q2) / r^2.
Step 4: Comparing the forces
- Comparing the new force F' with the given force phy/4, we have: F' = 16 * (k * q1 * q2) / r^2 = 16 * (phy/4) = 4phy.
Conclusion:
- The force acting between the charges, when the magnitudes are doubled and the distance is halved, is 4phy.
- Therefore, the correct option is (b) 4phy.
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The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy?
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The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy? 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 The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy?.
The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy? 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 The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The force acting between two point charges kept at a certain distance is phy/4. Now magnitudes of charges are doubled and distance between them is halved, the force acting between them is ... (a)phy (b)4 phy (c)8phy (d)16phy?.
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