Two charges $5 imes 10^{-8} ext{ C}$ and $-3 imes 10^{-8} ext{ C}$ are located $16 ext{ cm}$

Question

Two charges $5 	imes 10^{-8}	ext{ C}$ and $-3 	imes 10^{-8}	ext{ C}$ are located $16	ext{ cm}$ apart. At what point on the line segment joining the two charges is the electric potential zero?

Professor PC Thomas Classes · Answer

Let distance from the positive charge be $x$ cm. $V = k\frac{5 	imes 10^{-8}}{x} + k\frac{-3 	imes 10^{-8}}{16-x} = 0 \implies \frac{5}{x} = \frac{3}{16-x} \implies 80 - 5x = 3x \implies 8x = 80 \implies x = 10	ext{ cm}$.