Question

A point charge $+Q$ having mass $m$ is fixed on horizontal smooth surface. Another point charge having magnitude $+2Q$ and mass $2m$ is projected horizontal towards the charge $+Q$ from far distance with velocity $v_o$.

Acceleration of the particle $+2Q$ when it is closest to fixed particle $+Q$ :

• A. Zero
• B. $\frac{m{v}_o^4}{2k{Q}^2}$
• C. $\frac{m{v}_o^4}{4k{Q}^2}$
• D. $\frac{m{v}_o^4}{k{Q}^2}$

Answer 1

The correct option is C $\frac{m{v}_o^4}{4k{Q}^2}$


At closest approach, $+2Q$ charge will be at instant rest.


On applying the law of conservation of mechanical energy between initial and final state, we get:

$\frac{1}{2}\times 2m\times v_o^2+\frac{1}{2}\times m\times 0^2+\frac{k\left(2Q\right)\left(Q\right)}{\infty }=0+0+\frac{k\left(2Q\right)\left(Q\right)}{d}$

$\Rightarrow m{v}_o^2=\frac{2k{Q}^2}{d}$

$\Rightarrow d=\frac{2k{Q}^2}{m{v}_o^2}...\left(i\right)$

Now, force of repulsion on the charge $+2Q$ by $+Q$,

$F=\frac{k\left(2Q\right)\left(Q\right)}{d^2}$

So, deceleration,
$a=\frac{F}{2m}=\frac{2k{Q}^2}{2m{d}^2}$

$\Rightarrow a=\frac{2k{Q}^2}{2m}\times {\left(\frac{m{v}_o^2}{2k{Q}^2}\right)}^2=\frac{m{v}_o^4}{4k{Q}^2}$