Two point charges $q_{1}$ and $q_{2},$ of magnitude $+10^{-8} \; C$ and $-10^{-8} \; C,$ respectively,are placed $0.1 \; m$ apart. Calculate the electric fields at points $A, B$ and $C$ shown in the figure.

(N/A) The electric field vector $\mathbf{E}_{1A}$ at $A$ due to the positive charge $q_{1}$ points towards the right and has a magnitude:
$E_{1A} = \frac{(9 \times 10^{9} \; N m^{2} C^{-2}) \times (10^{-8} \; C)}{(0.05 \; m)^{2}} = 3.6 \times 10^{4} \; N C^{-1}$
The electric field vector $\mathbf{E}_{2A}$ at $A$ due to the negative charge $q_{2}$ also points towards the right and has the same magnitude. Hence,the magnitude of the total electric field $E_{A}$ at $A$ is:
$E_{A} = E_{1A} + E_{2A} = 7.2 \times 10^{4} \; N C^{-1}$
$\mathbf{E}_{A}$ is directed towards the right.
The electric field vector $\mathbf{E}_{1B}$ at $B$ due to the positive charge $q_{1}$ points towards the left and has a magnitude:
$E_{1B} = \frac{(9 \times 10^{9} \; N m^{2} C^{-2}) \times (10^{-8} \; C)}{(0.05 \; m)^{2}} = 3.6 \times 10^{4} \; N C^{-1}$
The electric field vector $\mathbf{E}_{2B}$ at $B$ due to the negative charge $q_{2}$ points towards the right and has a magnitude:
$E_{2B} = \frac{(9 \times 10^{9} \; N m^{2} C^{-2}) \times (10^{-8} \; C)}{(0.15 \; m)^{2}} = 4 \times 10^{3} \; N C^{-1}$
The magnitude of the total electric field at $B$ is:
$E_{B} = E_{1B} - E_{2B} = 3.2 \times 10^{4} \; N C^{-1}$
$\mathbf{E}_{B}$ is directed towards the left.
The magnitude of each electric field vector at point $C$,due to charges $q_{1}$ and $q_{2}$ is:
$E_{1C} = E_{2C} = \frac{(9 \times 10^{9} \; N m^{2} C^{-2}) \times (10^{-8} \; C)}{(0.10 \; m)^{2}} = 9 \times 10^{3} \; N C^{-1}$
The resultant of these two vectors is:
$E_{C} = E_{1C} \cos(60^{\circ}) + E_{2C} \cos(60^{\circ}) = 2 \times (9 \times 10^{3}) \times 0.5 = 9 \times 10^{3} \; N C^{-1}$
$\mathbf{E}_{C}$ points towards the right.