Answer:
A. [tex] E=13775510.2\frac{N}{C}[/tex]
B. Towards negative charge
Explanation:
Electric field satisfies the superposition principle, that is, the sum of the individual electric fields on a point is the net electric field. So, the electric field on the point is the sum of the electric field of both charges. Because both are pointing towards negative charge.
[tex]E=E_1+E_2 [/tex]
The individual electric fields are:
[tex]E_1=k\frac{q_1}{r^{2}}[/tex]
and:
[tex] E_2}-k\frac{q_2}{r^{2}}[/tex]
With q the charges, r the distance charge-point and k the constant [tex] [/tex]
So:
[tex]E=k\frac{q_1}{r^{2}}+k\frac{q_2}{r^{2}}=\frac{k}{r^2}(q_1+q_2)[/tex]
[tex]E=\frac{9.0\times10^{9}}{(0.042)^2}(5.8\times10^{-6}+(5.0\times10^{-6}))=55102040.83\frac{N}{C}=5.5\times10^{7}\frac{N}{C} [/tex]
The direction of the electric field is towards negative charge because superposition principle and both are pointing towards charge, so the sum should do it too.
