Question

A particle with charge 2q on the negative x axis and a second particle with charge 4q on the positive x axis are each a distance d from the origin. Where should a third particle with charge 6q be placed so that the magnitude of the electric field at the origin is zero? (Use any variable or symbol stated above as necessary.)

Answer 1

Answer:

$x = \sqrt{2 d^2}= \pm \sqrt{2}d$

So the charge can be placed at $x =-\sqrt{2} d$ from the origin.

Explanation:

For this case we have the situation described in the figure attached.

We want to find the net electric field at the origin, and the net electric filed is the sum of the individual electric fields given by:

$E_{net}= E_{2q} +E_{4q} +E_{6q}$

Since the components are just in the x axis we have are ust focused on the electric fields along the x axis given by:

$E_{neto}= E_{2q} -E_{4q} +E_{6q}=0$

We can assume that the charge 6q would be to the left of the origin. If we replace the formulas we got this:

$0= k \frac{2q}{d^2} - k \frac{4q}{d^2} + k \frac{6q}{x^2}$

And for this case x represent the distance between the origin and the hypothetical particle

We can take common factor kq and we got:

$0 = kq [\frac{1}{d^2} - \frac{4}{d^2} +\frac{6}{x^2}]$

We divide both sides by kq and we got:

$0 =\frac{1}{d^2} - \frac{4}{d^2} +\frac{6}{x^2}$

And then we can solve for x like this:

$-\frac{6}{x^2}= \frac{1}{d^2} -\frac{4}{d^2}$

$-\frac{6}{x^2}= -\frac{3d^2}{d^4} =-\frac{3}{d^2}$

$x^2 = \frac{6d^2}{3}$

$x = \sqrt{2 d^2}= \pm \sqrt{2}d$

So the charge can be placed at $x =-\sqrt{2} d$ from the origin.