If point A is between the two charges. And charge q1 is 5cm from A and q2 is 15cm from A on the opposite side, what’s the net Electric Field on A

The distance from charge $q_1$ to point A is $r_1 = 5 \text{ cm} = 0.05 \text{ m}$.
The distance from charge $q_2$ to point A is $r_2 = 15 \text{ cm} = 0.15 \text{ m}$.
Let the magnitude of both charges be $q$.
Coulomb's constant is $k = 8.987 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$.
The formula for the electric field due to a point charge is $E = \frac{kq}{r^2}$.

Assume $q_1$ and $q_2$ are both positive charges. Let $q_1$ be to the left of A and $q_2$ be to the right of A.

• The electric field $E_1$ due to $q_1$ at A points away from $q_1$ (to the right).
• The electric field $E_2$ due to $q_2$ at A points away from $q_2$ (to the left).

Since $E_1$ points to the right and $E_2$ points to the left, and $E_1$ is larger than $E_2$, the net electric field will point to the right.

The net electric field at point A is approximately $3.196 \times 10^{12} q \text{ N/C}$ and points towards the charge $q_2$ (or away from $q_1$). The exact numerical value depends on the magnitude of the charge $q$.