Two tiny spheres carrying charges $1.5 \; \mu C$ and $2.5 \; \mu C$ are located $30 \; cm$ apart. Find the potential and electric field:
$(a)$ at the mid-point of the line joining the two charges,and
$(b)$ at a point $10 \; cm$ from this midpoint in a plane normal to the line and passing through the mid-point.

(N/A) Let the charges be $q_{1} = 1.5 \; \mu C$ and $q_{2} = 2.5 \; \mu C$ placed at points $A$ and $B$ respectively. The distance between them is $d = 30 \; cm = 0.3 \; m$.
$(a)$ At the midpoint $O$:
Distance from each charge is $r = d/2 = 0.15 \; m$.
Potential $V = \frac{1}{4 \pi \epsilon_{0}} \left( \frac{q_{1}}{r} + \frac{q_{2}}{r} \right) = (9 \times 10^{9}) \times \frac{10^{-6}}{0.15} (1.5 + 2.5) = 60 \times 10^{9} \times 10^{-6} \times 4 = 2.4 \times 10^{5} \; V$.
Electric field $E = \left| \frac{1}{4 \pi \epsilon_{0}} \frac{q_{2}}{r^{2}} - \frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}}{r^{2}} \right| = \frac{9 \times 10^{9} \times 10^{-6}}{(0.15)^{2}} (2.5 - 1.5) = \frac{9 \times 10^{3}}{0.0225} (1) = 4 \times 10^{5} \; V/m$.
$(b)$ At point $Z$ ($10 \; cm$ from midpoint):
Distance $AZ = BZ = \sqrt{(0.15)^{2} + (0.1)^{2}} = \sqrt{0.0225 + 0.01} = \sqrt{0.0325} \approx 0.18 \; m$.
Potential $V = \frac{1}{4 \pi \epsilon_{0}} \left( \frac{q_{1}}{AZ} + \frac{q_{2}}{BZ} \right) = \frac{9 \times 10^{9} \times 10^{-6}}{0.18} (1.5 + 2.5) = 2.0 \times 10^{5} \; V$.
Electric field components: $E_{A} = \frac{9 \times 10^{9} \times 1.5 \times 10^{-6}}{0.0325} \approx 4.15 \times 10^{5} \; V/m$,$E_{B} = \frac{9 \times 10^{9} \times 2.5 \times 10^{-6}}{0.0325} \approx 6.92 \times 10^{5} \; V/m$.
Resultant $E = \sqrt{E_{A}^{2} + E_{B}^{2} + 2 E_{A} E_{B} \cos(2\theta)}$,where $\cos\theta = 0.1/0.18 = 5/9$. Resultant $E \approx 6.6 \times 10^{5} \; V/m$.