Two charges $\pm 10\; \mu C$ are placed $5.0\; mm$ apart. Determine the electric field at $(a)$ a point $P$ on the axis of the dipole $15\; cm$ away from its centre $O$ on the side of the positive charge,as shown in Figure $(a),$ and $(b)$ a point $Q, 15\; cm$ away from $O$ on a line passing through $O$ and normal to the axis of the dipole,as shown in Figure.

(N/A) Field at $P$ due to charge $+10\; \mu C$ is $E_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q}{(r-a)^2} = \frac{9 \times 10^9 \times 10^{-5}}{(0.15 - 0.0025)^2} \approx 4.13 \times 10^6\; N/C$ along $BP$.
Field at $P$ due to charge $-10\; \mu C$ is $E_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q}{(r+a)^2} = \frac{9 \times 10^9 \times 10^{-5}}{(0.15 + 0.0025)^2} \approx 3.86 \times 10^6\; N/C$ along $PA$.
The resultant electric field at $P$ is $E_P = E_1 - E_2 = 2.7 \times 10^5\; N/C$ along $BP$.
Using the dipole formula $E = \frac{2p}{4 \pi \varepsilon_0 r^3}$ where $p = q(2a) = 10^{-5} \times 0.005 = 5 \times 10^{-8}\; C\cdot m$,we get $E = \frac{2 \times 9 \times 10^9 \times 5 \times 10^{-8}}{(0.15)^3} = 2.67 \times 10^5\; N/C$.
$(b)$ Field at $Q$ due to charge $+10\; \mu C$ at $B$ is $E_B = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2+a^2} = \frac{9 \times 10^9 \times 10^{-5}}{(0.15)^2 + (0.0025)^2} \approx 3.99 \times 10^6\; N/C$.
Field at $Q$ due to charge $-10\; \mu C$ at $A$ is $E_A = 3.99 \times 10^6\; N/C$.
The resultant field $E_Q = 2 E_B \cos \theta = 2 E_B \frac{a}{\sqrt{r^2+a^2}} = 2 \times 3.99 \times 10^6 \times \frac{0.0025}{\sqrt{0.15^2 + 0.0025^2}} \approx 1.33 \times 10^5\; N/C$ along $BA$.
Using the dipole formula $E = \frac{p}{4 \pi \varepsilon_0 r^3} = \frac{9 \times 10^9 \times 5 \times 10^{-8}}{(0.15)^3} = 1.33 \times 10^5\; N/C$ opposite to the dipole moment.