- Home
- Standard 12
- Physics
1. Electric Charges and Fields
medium
Two charges, each equal to $q$, are kept at $x = -a$ and $x = a$ on the $x-$axis. A particle of mass $m$ and charge $q_0=\frac{q}{2}$ is placed at the origin. If charge $q_0$ is given a small displacement $(y < < a)$ along the $y-$axis, the net force acting on the particle is proportional to
A
$y$
B
$-y$
C
$\frac{1}{y}$
D
$-$$\;\frac{1}{y}$
(JEE MAIN-2013)
Solution
$\Rightarrow \mathrm{F}_{\mathrm{net}}=2 \mathrm{F} \cos \theta$
$F_{n e t}=\frac{2 k q\left(\frac{q}{2}\right)}{(\sqrt{y^{2}+a^{2}})^{2}} \cdot \frac{y}{\sqrt{y^{2}+a^{2}}}$
$F_{n e t}=\frac{2 k q\left(\frac{q}{2}\right) y}{\left(y^{2}+a^{2}\right)^{3 / 2}} \Rightarrow \frac{k q^{2} y}{a^{3}}$
So, $F \propto y$
Standard 12
Physics
