The point charges +4 μC and –6 μC are separated by a distance of 20 cm in air. At what point of the line joining the two charges is the electric potential zero?

Let, the point charges be given by q_A and q_B. 
Charge, q_A= $+4\times {10}^{-6} \mathrm{C}$
Charge, q_B= $-6\times {10}^{-6}\mathrm{C}$
Distance betwen charges, d=20 cm
Let, the point where the potential is 0 be O.
Assume, the distance of O from point charge A be x.
then, distance of O from point charge B will be (.2-x)

Since the electric potential is zero at point O therefore,
Potential at point O due to q_A=Potential at point O due to q_B 
$$\begin{gathered} \Rightarrow {\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}=0 \\ \frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }}\frac{4\times {10}^{-6}}{\mathrm{x}}=- \frac{1}{4{\mathrm{\pi \epsilon }}_{\circ }} \frac{-6\times {10}^{-6}}{0.2-\mathrm{x}} \\ \frac{4}{\mathrm{x}}=\frac{6}{0.2-\mathrm{x}} \\ 0.8-4\mathrm{x}=6\mathrm{x} \\ 0.8=10\mathrm{x} \\ \Rightarrow \mathrm{x}= 0.08 \mathrm{m} \end{gathered}$$

Therefore, the point O is at a distance of 0.08 m from charge placed at A.