Two point charges $+5 \mu \mathrm{C}$ and $-2 \mu \mathrm{C}$ are kept at a distance of 1 m in free

Question

Two point charges $+5 \mu \mathrm{C}$ and $-2 \mu \mathrm{C}$ are kept at a distance of 1 m in free space. The distance between the two zero potential points on the line joining the charges is\

Professor PC Thomas Classes · Answer

\Let the potential be zero at points P and Q respectively. $P$ is at a distance $x_{1}$ from $+5 \mu C$ and $Q$ is at distance $x_{2}$ from $-2 \mu C$ as shown in the figure above.\ $	herefore$ Potential at P is\ $\mathrm{V}_{\mathrm{p}}=\frac{5 	imes 10^{-6}}{4 \pi \varepsilon_{0} \mathrm{x}_{1}}-\frac{2 	imes 10^{-6}}{4 \pi \varepsilon_{0}\left(1-\mathrm{x}_{\mathrm{i}}ight)}$\ As $\mathrm{V}_{\mathrm{p}}=0$ $$ \begin{aligned} & 	herefore \frac{5 	imes 10^{-6}}{4 \pi \varepsilon_{0} \mathrm{x}_{1}}=\frac{2 	imes 10^{-6}}{4 \pi \varepsilon_{0}\left(1-\mathrm{x}_{1}ight)} \ & 5-5 \mathrm{x}_{1}=2 \mathrm{x}_{1} \ & 7 \mathrm{x}_{1}=5 \ & \mathrm{x}_{1}=\frac{5}{7} \end{aligned} $$ Potential at Q is\ $\mathrm{V}_{\mathrm{Q}}=\frac{5 	imes 10^{-6}}{4 \pi \varepsilon_{0}\left(1+\mathrm{x}_{2}ight)}-\frac{2 	imes 10^{-6}}{4 \pi \varepsilon_{0} \mathrm{x}_{2}}$\ $\mathrm{AsV}_{\mathrm{Q}}=0 	herefore \frac{5}{1+\mathrm{x}_{2}}=\frac{2}{\mathrm{x}_{2}}$\ $5 \mathrm{x}_{2}=2+2 \mathrm{x}_{2}$\ $3 \mathrm{x}_{2}=2$\ $\mathrm{x}_{2}=\frac{2}{3}$\ $	herefore$ Required distance between P and Q\ $=\left(1-\mathrm{x}_{1}ight)+\mathrm{x}_{2}$\ $=1-\frac{5}{7}+\frac{2}{3}=\frac{21-15+14}{21}$\