Question

In active example 19-1, the author finds the point on the axis connecting a -5.4 ?C charge and a -2.2?C charge where a +1.6?C charge would feel no net force. Where would this point be if we doubled the charge of the q₂ to -4.4?C (q₁ and q₃ remain as in the problem.) At the bottom of the answers is the picture the question is referring to.

	1.	0.61m
	2.	0.52m
	3.	0.70m
	4.	1.52m
	5.	none of the above is correct

Answer 1

If the charge q₂ is doubled while the distance between q₁ and q₂ is kept constant then the net Force on q₃ is given by:

$F_3 = \frac{1}{4\pi \epsilon_o}[-\frac{1.6\times 10^{-6}(5.4\times 10^{-6})}{x^2} + \frac{1.6\times 10^{-6}(4.4\times 10^{-6})}{(1-x)^2}]\hat{x}$

Substitute F₃ = 0 to get a value of x (from the origin where q₁ is positioned).

$\frac{1.6\times 10^{-6}(5.4\times 10^{-6})}{x^2} = \frac{1.6\times 10^{-6}(4.4\times 10^{-6})}{(1-x)^2}$

=> $0 = \frac{1}{4\pi \epsilon_o}[-\frac{1.6\times 10^{-6}(5.4\times 10^{-6})}{x^2} + \frac{1.6\times 10^{-6}(4.4\times 10^{-6})}{(1-x)^2}]$

=> $\frac{1.6\times 10^{-6}(5.4\times 10^{-6})}{x^2} = \frac{1.6\times 10^{-6}(4.4\times 10^{-6})}{(1-x)^2}$

=> 27 - 27x² = 22x²

=> x = 0.551 m

hence option 2.