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Question 10. Each of the two point charges are doubled and their distance is halved. Force of interaction becomes p times, where p is: (a) 1 (b) 4. (C) (d) 16 16 |
| Answer» C 16Explanation:By Coulomb's LAW, the force between TWO charges q 1 and q 2 is F= r 2 KQ 1 q 2 where r= separation between charges.When charges are doubled and their distance is halved , the force of interaction becomes F ′ = (r/2) 2 k(2q 1 )(2q 2 ) =16FThus, the value of n will be 16. | |