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 ?
Add an Answer
More Questions From This User See All
Answers & Comments
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:
mark me brainiest
hope this helps