Solution:
Two tiny spheres carrying charges are located at points $A$ and $B$
The charge at point $A, q_{1}=1.5 \mu \mathrm{C}$
The charge at point $B, q_{2}=2.5 \mu \mathrm{C}$
The distance between the two charges $=30 \mathrm{~cm}=0.3 \mathrm{~m}$
(a) Let 0 be the midpoint. Let $V_{1}$ and $E_{1}$ be the potential and electric field respectively at the midpoint. ${V}_{1}={P}$ otential due to charge at ${A}+{P}$ otential due to charge at ${B}$
$V_{1}=\frac{q_{1}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)}+\frac{q_{2}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)} V_{1}=\frac{1}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)}\left(q_{1}+q_{2}\right)$
Here,
$\varepsilon_{0}=P$ ermittivity of the free space
$$
\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} N C^{2} m^{-2}
$$
Therefore,
$V_{1}=\frac{9 \times 10^{9} \times 10^{-6}}{\left(\frac{0.30}{2}\right)}(2.5+1.5)=2.4 \times 10^{5} \mathrm{~V}$
Electric field at $0, E_{1}=E$ lectric field due to $q_{2}-$ Electric field due to $q_{1}$
$=\frac{q_{2}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)^{2}}-\frac{q_{1}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)^{2}}=\frac{9 \times 10^{9}}{\left(\frac{0.30}{2}\right)^{2}} \times 10^{-6} \times(2.5-1.5)$
$=400 \times 10^{3} \mathrm{~V} / \mathrm{m}$
Therefore, the potential at midpoint is $2.4 \times 10^{5} \mathrm{~V}$ and the electric field at the midpoint is $400 \times 10^{3} \mathrm{~V} / \mathrm{m}$.
(b) Consider a point $Z$ such that the distance $0 Z=10 \mathrm{~cm}=0.1 \mathrm{~m}$ as shown in the figure.
Let $V_2$ and $E_2$ represent the potential and electric field, respectively, at the point $Z$ in the coordinate system. The distance between the two points
$B Z=A Z=\sqrt{(0.1)^{2}+(0.15)^{2}}=0.18 \mathrm{~m}$
The potential at $V_{2}=P$ otential due to the charge at $A+P$ otential due to the charge at $B$
$=\frac{q_{1}}{4 \pi \epsilon_{0}(A Z)}+\frac{q_{2}}{4 \pi \epsilon_{0}(B Z)}$
$=\frac{9 \times 10^{9} \times 10^{-6}}{0.18}(1.5+$
$2.5)$
$=2 \times 10^{5} \mathrm{~V}$
The electric field due to $\mathrm{q}_{1}$ at $\mathrm{Z}$
$E_{A}=\frac{q_{1}}{4 \pi \epsilon_{0}(A Z)^{2}}=\frac{9 \times 10^{9} \times 1.5 \times 10^{-6}}{(0.18)^{2}}$
$=416 \times 10^{3} \mathrm{~V} / \mathrm{m}$
The electric field due to $q_{2}$ at $Z$
$E_{B}=\frac{q_{2}}{4 \pi \epsilon_{0}(B Z)^{2}}=\frac{9 \times 10^{9} \times 2.5 \times 10^{-6}}{(0.18)^{2}}$
$=694 \times 10^{3} \mathrm{~V} / \mathrm{m}$
The resultant field intensity at $\mathrm{Z}$ $E=\sqrt{E_{A}^{2}+E_{B}^{2}+2 E_{A} E_{B} \cos 2 \theta}$
From the figure we get $\cos \theta=(0.10 / 0.18)=5 / 9=0.5556$
$\theta=\cos ^{-1}(0.5556)=56.25$
$2 \theta=2 \times 56.25=112.5^{\circ}$
$\cos 2 \theta=-0.38$
$E=\sqrt{\left(0.416 \times 10^{6}\right)^{2}+\left(0.69 \times 10^{6}\right)^{2}+2 \times 0.416 \times 0.69 \times 10^{12} \times(-0.38)}$
$=6.6 \times 10^{5} \mathrm{~V} / \mathrm{m}$
Therefore the potential at the point $Z$ is $694 \times 10^{3} \mathrm{~V} / \mathrm{m}$ and the electric field is $6.6 \times 10^{5} \mathrm{~V} / \mathrm{m}$.