Two identical metal spheres carry charges of +q and -2q respectively. When the spheres are seperated by a large distance r , the force b/w them is F. Now the spheres are allowed to touch and then moved back to the same seperation.Find the new force of repulsion b/w them.

Question

Answer

F=2kq^2/r^2

Now after touching sphere charges become-

-2q+q/2=-q/2

Now new force-

F'=-kq^2/4r^2

So F'=F/8

Answer

Answer

F/8

Answer

Initial Setup:
- Two identical metal spheres with charges of q and -2q respectively
- Separated by a large distance r
- Force of repulsion between them is F

Explanation:
- Coulomb's law states that the force of repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
- Therefore, the force of repulsion between the two spheres can be calculated using the equation F = k * (q * (-2q)) / r^2, where k is the Coulomb constant.
- After the spheres are allowed to touch, they will share their charges and redistribute them evenly.
- This means that the final charges of the spheres will be -q and -q.
- When the spheres are moved back to the same separation distance, the force of repulsion between them can be calculated using the same equation as before.
- However, the charges of the spheres have changed, so the equation becomes F' = k * ((-q) * (-q)) / r^2.
- Simplifying this equation gives F' = k * (q^2) / r^2.
- Comparing this equation to the previous one, we see that the force of repulsion has decreased because the charges of the spheres are now equal and opposite instead of unequal.

Conclusion:
- The new force of repulsion between the spheres is F' = k * (q^2) / r^2, which is less than the initial force F.
- This occurs because the charges of the spheres are redistributed evenly after they touch, resulting in equal and opposite charges rather than unequal charges.

Answer

F/8

Answer

Please give me brief ans

Answer

Answer

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