Question

A particle, having a charge $+5\mu C$, is initially at rest at the point $x=30cm$ on the $x-$axis. The particle begins to move due to the presence of a charge $Q$ that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved $15cm$ from its initial position if
(a) $Q=+5\mu C$ and (b) $Q=-15\mu C$

Answer 1

Gain in K.E$=$Difference in P.E

Given:$q_1=5\mu$C, $q_2=\pm 15\mu$C and $r_{12}=30$

K.E at the instant $q_1$ moved $15$cm from the initial position.

Initial P.E, $E=\frac{9\times {10}^9\times 5\times {10}^{-6}\times 15\times {10}^{-6}}{3\times {10}^{-2}}$

$\left(a\right)$Since both charges are positive, the charges repel.Hence, the charge $q_1$ moves away by $15$cm.

Hence, $r_{12}=30+15=45$cm

$E^{\prime }=\frac{9\times {10}^9\times 5\times {10}^{-6}\times 15\times {10}^{-6}}{45\times {10}^{-2}}=1.5$

K.E$=E-E^{\prime }=2.25-1.5=0.75$

$\left(b\right)$Since both charges are opposite, the charges attract.Hence, the charge $q_1$ moves closer by $15$cm.

Hence, $r_{12}=30-15=15$cm

$E^{\prime }=\frac{9\times {10}^9\times 5\times {10}^{-6}\times 15\times {10}^{-6}}{15\times {10}^{-2}}=4.5$

K.E$=E^{\prime }-E=4.5-2.25=2.25$