# Question #3949e

Jun 2, 2017

The point charge A carries charge ${q}_{A} = 4 \mu C$

The point charge B carries charge ${q}_{B} = - 9 \mu C$

The distance between A and B is $0.5 m$

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 $y - x = 0.5 m \ldots \ldots \left[1\right]$

Again magnitude Intensity at neutral point due to A

${E}_{A} = k \times {q}_{A} / {x}^{2}$

And magnitude Intensity at neutral point due toB

${E}_{A} = k \times {q}_{B} / {y}^{2}$, where k =Couomb's constant

So

${E}_{A} = {E}_{B}$

$\implies k \times {q}_{A} / {x}^{2} = k \times {q}_{B} / {y}^{2}$

$\implies \frac{4 \times {10}^{-} 6}{x} ^ 2 = \frac{9 \times {10}^{-} 6}{y} ^ 2$

$\implies {y}^{2} / {x}^{2} = \frac{9}{4}$

$\implies \frac{y}{x} = \frac{3}{2}$

$\implies y = 1.5 x \ldots . . \left[2\right]$

From [1] and [2]

$0.5 + x = 1.5$

$\implies x = 1 m$

So $y = 1.5 m$