Answer:
[tex]E_{net} = 3.6 \times 10^4 N/C[/tex]
Explanation:
As the two charges Q1 and Q2 are placed at some distance apart
so the electric field at mid point will be twice the electric field due to one charge
Because here the two charges are of opposite sign so here the electric field at mid point will be added due to both
so here we have
[tex]E_{net} = 2E[/tex]
[tex]E_{net} = 2(\frac{kQ}{r^2})[/tex]
distance of mid point from one charge is given as
[tex]r = \frac{\sqrt{1^2 + 1^2}}{2}[/tex]
[tex]E_{net} = 2 (\frac{(9\times 10^9)(1\times 10^{-6})}{(\frac{1}{\sqrt2})^2}[/tex]
[tex]E_{net} = 3.6 \times 10^4 N/C[/tex]
