Grade 12th passElectrostatics

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...

Sheema

9 Years agoGrade 12th pass

4 Answers

Vikas TU

9 Years ago

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.

Akshat

8 Years ago

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

Adarsh

8 Years ago

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.

Rishi Sharma

6 Years ago

Dear Student,
Please find below the solution to your problem.

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
Hence the new force becomes=> F/8.

Thanks and Regards