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Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!!
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Two small conducting spheres of equal radius have charges +10 microC &...
Concept: Charge transfer from higher charge to lower charge due to contact, ie. +10 to -20. Step 1: Initial force = K(+10×-20)÷r^2Step 2: New charge = (-20+10)÷2=-5 (Mean of total charge).Step 3: New force = K(-5×-5)÷r^2(Distance unchanged as the charges were brought back after contact)Step 4: Ratio(F1:F2)= -200:25 = -8:1 Hope it helps....
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Two small conducting spheres of equal radius have charges +10 microC &...
Understanding the Problem
Two small conducting spheres with charges +10 μC and -20 μC are placed at a distance R apart. Initially, they experience a force F1 due to their charges. When brought into contact, they share their charges evenly, and then when separated back to the distance R, they experience a new force F2. We need to find the ratio F1:F2.
Calculating F1
- The force between two charged spheres is given by Coulomb's Law:
\[ F = k \frac{|q_1 \cdot q_2|}{R^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges.
- For initial charges:
\( q_1 = +10 \, \mu C \)
\( q_2 = -20 \, \mu C \)
\[ F1 = k \frac{|10 \cdot (-20)| \times 10^{-12}}{R^2} = k \frac{200 \times 10^{-12}}{R^2} \]
Calculating F2
- When the spheres are brought in contact, they share charges:
Total charge = \( +10 \, \mu C - 20 \, \mu C = -10 \, \mu C \)
Each sphere will have:
\[ q' = \frac{-10 \, \mu C}{2} = -5 \, \mu C \]
- Now, the force F2 when separated back to distance R:
\[ F2 = k \frac{|-5 \cdot -5| \times 10^{-12}}{R^2} = k \frac{25 \times 10^{-12}}{R^2} \]
Finding the Ratio F1:F2
- We can now find the ratio:
\[ \frac{F1}{F2} = \frac{k \frac{200 \times 10^{-12}}{R^2}}{k \frac{25 \times 10^{-12}}{R^2}} = \frac{200}{25} = 8 \]
- Thus, the ratio \( F1 : F2 = 8 : 1 \).
Conclusion
The ratio of forces F1 to F2 is \( 8 : 1 \).
Two small conducting spheres with charges +10 μC and -20 μC are placed at a distance R apart. Initially, they experience a force F1 due to their charges. When brought into contact, they share their charges evenly, and then when separated back to the distance R, they experience a new force F2. We need to find the ratio F1:F2.
Calculating F1
- The force between two charged spheres is given by Coulomb's Law:
\[ F = k \frac{|q_1 \cdot q_2|}{R^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges.
- For initial charges:
\( q_1 = +10 \, \mu C \)
\( q_2 = -20 \, \mu C \)
\[ F1 = k \frac{|10 \cdot (-20)| \times 10^{-12}}{R^2} = k \frac{200 \times 10^{-12}}{R^2} \]
Calculating F2
- When the spheres are brought in contact, they share charges:
Total charge = \( +10 \, \mu C - 20 \, \mu C = -10 \, \mu C \)
Each sphere will have:
\[ q' = \frac{-10 \, \mu C}{2} = -5 \, \mu C \]
- Now, the force F2 when separated back to distance R:
\[ F2 = k \frac{|-5 \cdot -5| \times 10^{-12}}{R^2} = k \frac{25 \times 10^{-12}}{R^2} \]
Finding the Ratio F1:F2
- We can now find the ratio:
\[ \frac{F1}{F2} = \frac{k \frac{200 \times 10^{-12}}{R^2}}{k \frac{25 \times 10^{-12}}{R^2}} = \frac{200}{25} = 8 \]
- Thus, the ratio \( F1 : F2 = 8 : 1 \).
Conclusion
The ratio of forces F1 to F2 is \( 8 : 1 \).
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Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!!
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Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!! for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!! covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!!.
Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!! for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!! covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two small conducting spheres of equal radius have charges +10 microC & -20microC respectively placed at distance R frm each other. They experience force F1 . If they r brought in contact & separated to the same distance, they experience force F2..The ratio of F1 to F2 : 1:8-8:1 1:2-2:1# plz expln...!!!!.
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