Question

1.
1
Two equal charges are initially kept a certain distance apart in vacuum. If the magnitude
of each charge is increases by 50% the distance between them is increased by 100% and
the charge were kept in a medium of dielectric constant 9, the magnitude of the coulomb
force between them will charge :
a) % of initial value b) %16 of initial value
c) 14 of initial value d) 132 of initial value

Answer 1

Explanation:

Let say for simplicity two equal charges be of 1 coulmb of each.And distance b/w these be 1 m

So by the coulmb law, force b/w these two charges in vaccume would be F = 1 N.....(1)

If charges increased by 50% each mean each becomes 1.5 of magnitude.Also after increasing the distance of 100% meance double the distance between them.i.e 2 m

So,now again by coulumb law force b/w these two will be F' = (9*2.25)/4= 5.0625 N

Which means change in force= 4.0625 N

which is 4.0625 times of initial force of 1 N

Hence percentage change 406.25 %. <= Ans

I hope ,It was helpful to you.