Answer:
[tex]E_{net} = 1.44\times 10^{7}\,\frac{N}{C}[/tex]
Explanation:
Let consider that electric field lines from positive charge (element 1) moves away in a radial direction, whereas lines from negative charge (element 2) get closer in radial direction. Besides, let assume that the positive charge (element 1) is located on the right and the negative charge (element 2) is located on the left.
Lastly, let consider that any vector facing right has a positive direction and any vector facing left has a negative direction. The net electric field is obtained by applying the Principle of Superposition:
[tex]E_{net} = -E_{1}-E_{2}[/tex]
[tex]E_{net} = -\frac{2\cdot k\cdot q}{r^{2}}[/tex]
[tex]E_{net} = -\frac{2\cdot (9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (2\times 10^{-6}\,C)}{(0.05\,m)^{2}}[/tex]
[tex]E_{net} = 1.44\times 10^{7}\,\frac{N}{C}[/tex]
