Two charges $3 imes 10^{-8} ext{ C}$ and $-2 imes 10^{-8} ext{ C}$ are located $15 ext{ cm}$

Question

Two charges $3 	imes 10^{-8}	ext{ C}$ and $-2 	imes 10^{-8}	ext{ C}$ are located $15	ext{ cm}$ apart. At what point on the extended line joining the charges (on the side of the negative charge) is the electric potential zero?

Professor PC Thomas Classes · Answer

Let the point be at distance $x$ from the positive charge. Since it's on the side of the negative charge, the distance to the negative charge is $x - 15$. $V = k\frac{3 	imes 10^{-8}}{x} + k\frac{-2 	imes 10^{-8}}{x - 15} = 0 \Rightarrow \frac{3}{x} = \frac{2}{x - 15} \Rightarrow 3x - 45 = 2x \Rightarrow x = 45	ext{ cm}$.