Two charges4q and q are placed 30 cm apart at what point the value of electric force on the test charge will be zero

Where E=0 this point is called neutral point.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge 
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is 
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.