Question

Two small, identical in size, charged metallic spheres A and B repel each other with a force F when placed at a distance r between their centres. The spheres A and B are touched by identical, but uncharged ,spheres C and D respectively which are then removed away. The sphere B is now brought closer to A to a distance r/2 between their centres. What is the new forced of repulsion between A and B ?

Answer 1

Answer:

Explanation:

Let the original force b/w A & B be =

F=kq1q2/r^2

As A and C have same sizes, charges are shared equally.

Again as B and D have same sizes,their charges are also shared equally.

As charges on A and B are halved and acc. to question. distance is also halved from r to r/2.

So, F'=k(q1/2)(q2/2)/(r/2)^2

F' =kq1q2/r^2. =F

Hence , the force remains same.

Hope it helps .

Answer 2

Answer:

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hope this helps