Two charges $q_1=4 \times 10^{-7} C$ and $q_2=-6 \times 10^{-7} C$ are located 3 m apart. At what point the line joining the two charges is the electric potential zero? (Consider potential zero at infinity.)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.2 meter from charge $q_1$
The electric potential V due to a point charge q.
$V=\frac{kq}{r}$
and,
$V_1+V_2=0$
$∴\frac{kq_1}{x}+\frac{kq_2}{3-x}=0$
$\frac{4×10^{-7}}{x}+\frac{-6×10^{-7}}{3-x}=0$
$\frac{4}{x}-\frac{6}{3-x}=0$
$x=\frac{12}{10}=1.2m$