q1 and q2, if the electric field one fourth of the way from q1 to q2 is zero?
The answer is that the ratio of q1 and q2 is 1:9. How did they get this? Did the answer relate, somehow to this equation E= k.q1.q2/r^2 ?
Thanks!
2 Answers

Absolutely.
Consider 'r' as the distance between the two charges q1 and q2.
The force experienced by a unit charge (+ coulomb) at point P due to q1 is k.q1/d^2 .
Let the point P be at a distance of 'r'/4 from q1 along the line joining q1 and q2. Then it will be at a distance of 3r/4 from q2.
Hence 'F1' = k.16.q1.(1)/r^2.
Similarly the unit charge will experience a force due to q2 and it is 'F2' = k.16.q2.(1)/9.r^2 ?
For the point P to not experience any displacement due to the forces 'F1' and 'F2' they have to be equal an opposite.
k.16.q1.(1)/r^2. = k.16.q2.(1)/9.r^2
q1/q2 = 1/9.
Charges are Like that is both are +ve or ve.

E1 = E2
k * q1/x^2 = k * q2/(r  x)^2
> x = (1/4)r
q1/q2 = (x/r  x)^2
= (1/4 r/3/4 r)^2
= (1/3)^
> q1/q2 = 1/9
> q1 : q2 = 1 : 9