What third charge should be placed at x=+26cm so that the total electric field at x=+13.0cm is zero?
Express your answer to three significant figures and include appropriate units"
Let's call A the point at x=+13 cm where the total electric field is zero.
First of all, we can calculate the total field generated by the two charges [tex]q_1=2.20~pC=2.2\cdot10^{-9}~C[/tex] and [tex]q_2=-4.6~pC=4.6\cdot10^{-9}~C[/tex]. The two charges have a distance from point A of [tex]r_1=13~cm=0.13~m[/tex] and [tex]r_2=26~cm=0.26~m[/tex], respectively. The field generated by the positive charge heads towards right in point A, while the one generated by the negative charge heads towards left, so we should consider a sign - on this field. Therefore, the total field generated by charge 1 and 2 in A is
[tex]E_{12}=k_e \frac{q_1}{r_1^2}-k_e \frac{q_2}{r_2^2}=1170.3~V/m - 611.7~V/m=558.6~V/m [/tex]
In order to have total net field of zero in point A, the field generated by charge 3 should be equal to this value, and should point towards the left, so it must be a positive charge. So, we have:
[tex]E_3 = E_{12}=558.6~V/m=k_e \frac{q_3}{r_3^2} [/tex]
where [tex]r_3=0.13~m[/tex] is the distance of the charge 3 from point A. So we find
[tex]q_3=E_{12} \frac{r_3^2}{k_e}=1.05\cdot 10^-9~C=1.05~pC [/tex]
