Q. Two charges $+ 6\ \mu C$ and $+ 15\ \mu C$ are placed along the $x -$axis at $x = 0$ and $x = 2$m respectively A negative charge is placed between them such that the resultant force on it is zero. The negative charge is placed at
A) $x = 0.775m$
B) $x = 1.2m$
C) $x = 0.5m$
D) Position depends on the amount of charge
Explanation
(a) [image:1_image273.emf]
Let the charge $( - q)$ be placed at a distance $x$ from $+ 6\mu C.$ As the resultant force on $( - q)$ is zero
$\therefore\ \frac{( - q) \times 6 \times 10^{- 6}}{4\pi\varepsilon_{0} \times x^{2}} = \frac{( - q) \times 15 \times 10^{- 6}}{4\pi\varepsilon_{0} \times (2 - x)^{2}}$
$\frac{6}{x^{2}} = \frac{15}{(2 - x)^{2}} \Rightarrow \frac{1}{x} = \pm \frac{\sqrt{5}}{\sqrt{2}(2 - x)}$
On solving, we get $x = 0.775m$ or $x = \ - 3.44m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.