Question

Two charge particle P & Q having same charge $1\mu C$ and mass $4\mu kg$ are initially kept at the distance of $1mm$. Charge P is fixed, then the velocity of charge particle Q when the separation between them becomes $9mm$.

• A. $3\times {10}^{3}m/s$
• B. $2\times {10}^{3}m/s$
• C. $5\times {10}^{3}m/s$
• D. $7\times {10}^{3}m/s$

Answer 1

Answer: (B)

$2\times {10}^{3}m/s$

Loss in $PE=$ gain in $KE$
$K\times {10}^{-6}\times {10}^{-6}[\frac{1}{{10}^{-3}}-\frac{1}{9\times {10}^{-2}}]=\frac{1}{2}m{v}^2$
$9\times {10}^9\times \frac{{10}^{-6}\times {10}^{-6}}{{10}^{-3}}\times \frac{8}{9}=\frac{1}{2}\times 4\times {10}^{-6}\times V^2$
$V=2\times {10}^{3}m/s$