# Force between 2 charges increased by 16 times their distance =?

A. 4

B. 2

C. 1/8

D. 1/4

E. 8

Update:

times of original distance

times of original distance

• Anonymous
6 days ago

Force F, between 2 charges q1 and q2 with distance r =

F = K*q1*q2/r²

=> r = √(K*q1*q2/F)

When Force increased by 16 times ; F becomes 16F

r’ = √(K*q1*q2/16 F) = (1/4) * √(K*q1*q2/F) = (1/4)r

Their distance decreased by 4 times of original distance

Correct= D. 1/4

• ?
6 days ago

D.1/4

• Anonymous
6 days ago

D.1/4

• ?
6 days ago

Distance indirectly proportional to root of Force.

So Distance= root(1/16) =1/4

D is correct.

• Mr.Mukherjee
6 days ago

inversely proportional square root of the force 16 is the distance between the two charge particles, that is 1/4

• OMPRAKASH
6 days ago

we know from coulomb’s law that F inversely proportional to square of distance r . So if F1 ,F2 and r1,r2 are the initial forces and distances then F1/F2=r2^/r1^ i.e 1/16=r2^/r1^ i.e r2=r1/4 so distance decreased by1/4 times. so answer D is correct

• Hamilton
6 days ago

its like my third grade teacher’s stepfather’s nephew use to say, “don’t fiddle my oat bran.”

• tom7railway
6 days ago

Write the question out properly please, it still doesn’t make sense.