Two point charges $q_{A} = 3\; \mu C$ and $q_{B} = -3\; \mu C$ are located $20\; cm$ apart in vacuum.
$(a)$ What is the electric field at the midpoint $O$ of the line $AB$ joining the two charges?
$(b)$ If a negative test charge of magnitude $1.5 \times 10^{-9}\; C$ is placed at this point,what is the force experienced by the test charge?

(N/A) The situation is represented in the figure. $O$ is the mid-point of line $AB.$
Distance between the two charges,$AB = 20\; cm = 0.2\; m$.
$\therefore AO = OB = 10\; cm = 0.1\; m$.
Electric field at point $O$ due to charge $q_{A}$ $(+3\; \mu C)$:
$E_{A} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{|q_{A}|}{AO^{2}} = 9 \times 10^{9} \times \frac{3 \times 10^{-6}}{(0.1)^{2}} = 2.7 \times 10^{6}\; N/C$ (directed from $A$ to $B$).
Electric field at point $O$ due to charge $q_{B}$ $(-3\; \mu C)$:
$E_{B} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{|q_{B}|}{OB^{2}} = 9 \times 10^{9} \times \frac{3 \times 10^{-6}}{(0.1)^{2}} = 2.7 \times 10^{6}\; N/C$ (directed from $A$ to $B$).
Since both fields are in the same direction,the net electric field $E = E_{A} + E_{B} = 5.4 \times 10^{6}\; N/C$ directed along $OB$.
$(b)$ Force $F$ on a test charge $q = -1.5 \times 10^{-9}\; C$ placed at $O$:
$F = qE = (-1.5 \times 10^{-9}) \times (5.4 \times 10^{6}) = -8.1 \times 10^{-3}\; N$.
The negative sign indicates that the force is directed opposite to the electric field,i.e.,along $OA$.