$(a)$ Field at $P$ due to charge $+10\, \mu \,C$

$=\frac{10^{-5}\, C }{4 \pi\left(8.854 \times 10^{-12} \,C ^{2}\, N ^{-1}\, m ^{-2}\right)} \times \frac{1}{(15-0.25)^{2} \times 10^{-4} \,m ^{2}}$

$=4.13 \times 10^{6} \,N\, C ^{-1}$ along $BP$

Field at $P$ due to charge $-10 \,\mu\, C$

$=\frac{10^{-5} \,C }{4 \pi\left(8.854 \times 10^{-12} \,C ^{2} \,N ^{-1}\, m ^{-2}\right)} \times \frac{1}{(15+0.25)^{2} \times 10^{-4} \,m ^{2}}$

$=3.86 \times 10^{6} \,N\, C ^{-1}$ along $PA$

The resultant electric field at $P$ due to the two charges at $A$ and $B$ is $=2.7 \times 10^{5} \,N\, C ^{-1}$ along $BP$.

In this example, the ratio $OP/OB$ is quite large $(=60$ ). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges $\pm q,$ $2a$ distance apart, the electric field at a distance $r$ from the centre on the axis of the dipole has a magnitude

$E=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}} \quad(r / a \,>\, \,>\, 1)$

where $p=2 a q$ is the magnitude of the dipole moment. The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from $-q$ to $q$ ). Here, $p=10^{-5} \,C \times 5 \times 10^{-3} \,m =5 \times 10^{-8} \,C\,m$

Therefore,

$E=\frac{2 \times 5 \times 10^{-8} \,C\,m }{4 \pi\left(8.854 \times 10^{-12} \,C ^{2} \,N ^{-1}\, m ^{-2}\right)} \times \frac{1}{(15)^{3} \times 10^{-6}\, m ^{3}}$$=2.6 \times 10^{5}\, N\, C ^{-1}$

along the dipole moment direction $AB$, which is close to the result obtained earlier.

$(b)$ Field at $Q$ due to charge $+10\, \mu \,C$ at $B$

$=\frac{10^{-5} \,C }{4 \pi\left(8.854 \times 10^{-12}\, C ^{2} \,N ^{-1} \,m ^{-2}\right)} \times \frac{1}{\left[15^{2}+(0.25)^{2}\right] \times 10^{-4} \,m ^{2}}$

$=3.99 \times 10^{6}\, N\, C ^{-1}$ along $B Q$

Field at $Q$ due to charge $-10\, \mu \,C$ at $A$ $=\frac{10^{-5} \,C }{4 \pi\left(8.854 \times 10^{-12} \,C ^{2} \,N ^{-1} m ^{-2}\right)} \times \frac{1}{\left[15^{2}+(0.25)^{2}\right] \times 10^{-4} \,m ^{2}}$

$=3.99 \times 10^{6}\, N \,C ^{-1}$ along $QA$

Clearly, the components of these two forces with equal magnitudes cancel along the direction $OQ$ but add up along the direction parallel to $BA.$ Therefore, the resultant electric fleld at $Q$ due to the two charges at $A$ and $B$ is $=2 \times \frac{0.25}{\sqrt{15^{2}+(0.25)^{2}}} \times 3.99 \times 10^{6} \,N\, C ^{-1}$ along $BA$

$=1.33 \times 10^{5}\, N\, C ^{-1}$ along $BA$.

As in $(a)$, we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole

$E=\frac{p}{4 \pi \varepsilon_{0} r^{3}} \quad(r / a\,>\,>\,1)$

$=\frac{5 \times 10^{-8} \,C\,m }{4 \pi\left(8.854 \times 10^{-12} \,C ^{2}\, N ^{-1} \,m ^{-2}\right)} \times \frac{1}{(15)^{3} \times 10^{-6} \,m ^{3}}$

$=1.33 \times 10^{5} \,N\, C ^{-1}$

The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again, the result agrees with that obtained before.