91Ó°ÊÓ

$\mathit{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.0}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}$or${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{2}\mathit{e}$,${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{+}\mathbf{4}\mathit{e}\mathbf{,}$${\mathit{q}}_{\mathbf{3}}\mathbf{=}\mathbf{+}\mathit{e}\mathbf{,}$${\mathit{q}}_{\mathbf{4}}\mathbf{=}\mathbf{+}\mathbf{4}\mathit{e}\mathbf{,}$${\mathit{q}}_{\mathbf{5}}\mathbf{=}\mathbf{+}\mathbf{2}\mathit{e}$,${\mathit{q}}_{\mathbf{6}}\mathbf{=}\mathbf{+}\mathbf{8}\mathit{e}\mathbf{,}$${\mathit{q}}_{\mathbf{7}}\mathbf{=}\mathbf{+}\mathbf{6}\mathit{e}\mathbf{,}$$\mathbf{2}\mathit{d}$

Use Coulomb’s law to solve the problem.

As result of the fact that the Coulomb force is inversely proportional to$r^2$, a particle of charge Q that is distance d from the origin will exert a force on some charge qo at the origin of equal strength as a particle of charge 4Q at distance 2d would exert on$q_o$.

Therefore,${\mathit{q}}_{\mathbf{6}}\mathbf{=}\mathbf{+}\mathbf{8}\mathit{e}$on the –y axis could be replaced with a ‘+2e’ closer to the origin (at half the distance); this would add to the${\mathit{q}}_{\mathbf{5}}\mathbf{=}\mathbf{+}\mathbf{2}\mathit{e}$already there and produce ‘+4e’ below the origin, which exactly cancels the force due to$q_2=+4e$above the origin.

Similarly,${\mathit{q}}_{\mathbf{4}}\mathbf{=}\mathbf{+}\mathbf{4}\mathit{e}$to the far right could be replaced by a ‘+e’ at half the distance, which would add to${\mathit{q}}_{\mathbf{3}}\mathbf{=}\mathbf{+}\mathit{e}$already there to produce a ‘+2e’ at distance d to the right of the central charge${\mathit{q}}_{\mathbf{7}}$.

The horizontal force due to this ‘+2e’ is cancelled exactly by that of ${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{2}\mathit{e}$on the –x axis, so that the net force on$q_7$ is zero.