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Grade: 12th pass
        Two identical metal balls with charges + Q and - Q are separated by some distance and exert a force F on each other .they are joined by a conducting wire  which is then removed . the force between them will now be...
2 years ago

Answers : (3)

Vikas TU
9795 Points
							
Already answered ypur question,
Agn reposting for the ease of finding it as:
 
Before connecting a wire the force developed is:
F = 2KQ^2/r^2 (attracting)
AFter conecting the wire, the charge gets distributed equally hence,
the charge on each ball becmoes => (2Q – Q)/2 => Q/2.
The new Force b/w them is:
F’ = KQ^2/4r^2
Dividing both of them we get,
F/F’ = 8/1
Henc  ethe new force becomes=> F/8.
2 years ago
Akshat
12 Points
							Before connecting a wire the force developed is:F = 2KQ^2/r^2 (attracting)AFter conecting the wire, the charge gets distributed equally hence,the charge on each ball becmoes => (2Q – Q)/2 => Q/2.The new Force b/w them is:F’ = KQ^2/4r^2Dividing both of them we get,F/F’ = 8/1Henc  ethe new force becomes=> F/8
						
2 years ago
Adarsh
763 Points
							
Dear,
If two charge are joined with a conducting wire then transfer of charge will occur and it will stop when both potential will become equal .here, in this case their potential become zero and there will be no force to each other after joining wire.
 
2 years ago
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