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Question Number 131199 by aurpeyz last updated on 02/Feb/21

	Answered by physicstutes last updated on 02/Feb/21

	$Example 4$ $Q_{1} = 2.0 μ C and Q_{2} = - 4.0 μ C, R = 50 cm$ $the neutral point lies a distance of x from Q_{1} and (0.5 - x) m from Q_{2}$ $We define the electric field strenght as E = \frac{F}{Q}$ $E_{Q_{1}} = \frac{{k Q}_{2}}{R_{1}^{2}} = \frac{k (4.0 μ C)}{x^{2}}$ $E_{Q_{2}} = \frac{{k Q}_{1}}{R_{2}^{2}} = \frac{k (2.0 μ C)}{{(0.5 - x)}^{2}}$ $at the neutral point E_{Q_{1}} = E_{Q_{2}}$ $\Rightarrow \frac{4.0}{x^{2}} = \frac{2.0}{{(0.5 - x)}^{2}} \Rightarrow \frac{2}{x} = \frac{\sqrt{2}}{0.5 - x}$ $1 - 2 x = \sqrt{2} x$ $\Rightarrow x = \frac{1}{2 + \sqrt{2}} \approx 0.3 m or 30 cm from Q_{1}$
	Commented by aurpeyz last updated on 06/Feb/21

	$P l s c a n y o u e x p l a i n t h e n e u t r a l p o i n t$
	Commented by physicstutes last updated on 06/Feb/21

	$point were E_{n e t} = 0$
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