Question

Question #3949e

Answer 1

The point charge A carries charge `q_A=4muC`

The point charge B carries charge `q_B=-9muC`

The distance between A and B is `0.5m`

The electric intensity cannot be zero in between A and B as they are oppositely charged. The neutral point will be situated near A, the point charge of less magnitude and away from B , the point charge of greater magnitude,

Given that the intensity of the electric field is zero at x meters from A and y meters from B.

So `y-x=0.5m......[1]`

Again magnitude Intensity at neutral point due to A

`E_A=kxxq_A/x^2`

And magnitude Intensity at neutral point due toB

`E_A=kxxq_B/y^2`, where k =Couomb's constant

So

`E_A=E_B`

`=>kxxq_A/x^2=kxxq_B/y^2`

`=>(4xx10^-6)/x^2=(9xx10^-6)/y^2`

`=>y^2/x^2=9/4`

`=>y/x=3/2`

`=>y=1.5x.....[2]`

From [1] and [2]

`0.5+x=1.5`

`=>x=1m`

So `y=1.5m`