Question

Electric Field and Electric Field Lines
1. The point charges of q and 4q are kept 30 cm apart. At a distance on the straigh
line joining them, intensity of electric field is zero.
(A) 20 cm from 4q
(B) 7.5 cm from a
(C) 15 cm from 4g
(D) 5 cm from a​

Answer 1

Answer

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.

i hope u got ur answer.

Hope it helps