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...!!!!

Question

Answer

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....

Answer

-8:1

Answer

-8:1

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So F1/F2= -200/25= 8/1

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-8:1

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New charge=( Total charge*R1)/ R1+R2So new charge= -5

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8:1

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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 $.

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