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Q. A particle of mass 2g and charge $1 \mu C$ is held at a distance of 1m from a fixed charge $1 \mu C$. If the particle is released it will be repelled. The speed of particle when it is at a distance of 10 metre from the fixed charge is

Electric Charges and Fields

Solution:

Potential at $1\, m$ from the charge
$V _{ A }=\frac{ K \times 10^{-3}}{1}$
Potential at $10\, m$ from the charge
$V _{ B }=\frac{ K \times 10^{-3}}{10}= K \times 10^{-4}$
potential diff $= V _{ A }- V _{ B }$
$= K \left(10^{-3}-10^{-4}\right)$
Its velocity at $10 \,m$ is $V$, then
$\frac{1}{2} \times mv ^{2}=\left( V _{ A }- V _{ B }\right) q $
$\frac{1}{2} \times 2 \times 10^{-3} v ^{2}$
$= K \times 10^{-6}\left(1-\frac{1}{10}\right) \times 10^{-3}$
$v ^{2}= K \times \frac{9}{10} \times 10^{-6} $
$9 \times 10^{9} \times \frac{9}{10} \times 10^{-6}$
$=81 \times 100 $
$v =90 \,m / \sec$