Question

Two particles have equal masses of $5.0$g each and opposite charges of $+4.0\times {10}^{-5}$C and $-4.0\times {10}^{-5}$C. They are released from rest with a separation of $1.0$m between them. Find the speeds of the particles when the separation is reduced to $50$cm.

Answer 1

The charge on two particles is $q_1=q_2=4\times {10}^{-5}$

The distance between particles is $s=1m$

The mass of each particle is $m=5g=0.005$kg

The electrostatic force is given by:

$F=K\frac{q^2}{r^2}$

$=\frac{9\times {10}^9\times (4\times {10}^{-5}{)}^2}{1^2}=14.4$N

The acceleration 'a' of the particles will be:

$a=\frac{F}{m}$

$=\frac{14.4}{0.005}=2880m/s^2$

Now, $u=0$, $s=50cm=0.5m$,
$a=2880m/s^2$, $v=?$
The final speed of the particles can be obtained as:

$v^2=u^2+2as$
$\Rightarrow v^2=2\times 2880\times 0.5$

$\Rightarrow v=\sqrt{2880}=53.66$ m/s$\approx 54$ m/s for each particle.