A$+2.0\mathrm{nC}$charge is at the origin and a$-4.0\mathrm{nC}$charge is at$\mathrm{x}=1.0\mathrm{cm}$.

a. At what$\mathrm{x}$-coordinate could you place a proton so that it would experience no net force?

b. Would the net force be zero for an electron placed at the same position? Explain.

(a).

Figure is,

(a).

$+2\mathrm{nC}$at A causes a force:

${\mathrm{F}}_1=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{\left(2\mathrm{nc}\right)\mathrm{P}}{{\left(\mathrm{x}\times {10}^{-2}\right)}^2}$

Force resulting from$-4\mathrm{nC}$at A:

${\mathrm{F}}_2=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{\left(4\mathrm{nc}\right)\mathrm{P}}{{\left.(\mathrm{x}-1)\times {10}^{-2}\right)}^2}$

At${\mathrm{PF}}_{\text{net}}=0$,

${\mathrm{F}}_1={\mathrm{F}}_2$

localid="1649573924219" role="math" $\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{\left(2\mathrm{nc}\right)\mathrm{P}}{{\left(x\times {10}^{-2}\right)}^2}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{(-4\mathrm{nc})\mathrm{P}}{{\left((x-1)\times {10}^{-2}\right)}^2}$

localid="1649573963479" role="math" $$\begin{gathered} \frac{1}{x^2}=\frac{2}{(x-1{)}^2} \\ (x-1{)}^2=2x^2 \\ {x}^2-2x+1=2x^2 \\ {x}^2+2x-1=0 \end{gathered}$$

solve the above quadratic equation by quadratic equation, we get

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$where, role="math" localid="1649574059895" $a=1,b=2,c=-1$

role="math" localid="1649574190259" $$\begin{gathered} \Rightarrow x=\frac{-2\pm \sqrt{4-(4\times 1\times (-1\left)\right)}}{2\left(1\right)} \\ \Rightarrow x=\frac{-2\pm \sqrt{4+4}}{2} \\ \Rightarrow x=\frac{-2\pm \sqrt{8}}{2} \\ \Rightarrow x=\frac{-2\pm 2\sqrt{2}}{2} \\ \Rightarrow x=-1\pm \sqrt{2} \end{gathered}$$

$\therefore$The two real roots are

$x=-1-1.414=-2.414cm;x=-1+1.414=0.414cm$

From the above solution, the x-coordinate in which the proton placed is$-2.4cm$

(b). we need to verify the net force to be zero. $\therefore$Forces acting on electrons must be equal.

Force is,

${\mathrm{F}}_1={\mathrm{F}}_2$

localid="1649575589767" role="math" $$\begin{gathered} \Rightarrow \frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{\left(2\mathrm{n}\right)\mathrm{e}}{{(-2.4)}^2}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{\left(4\mathrm{n}\right)\mathrm{e}}{(-3.4{)}^2} \\ \Rightarrow \frac{0.346ne}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}=\frac{0.346ne}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}} \end{gathered}$$

$\therefore$The two forces acting on electrons are equal.

Yes. The net force between electrons within the same place is zero.