Question

A particle (charge = 5.0 μC) is released from rest at a point x = 10 cm on the x-axis. If a 5.0-μC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 90 cm?

Answer 1

Answer:

Explanation:

Given

Two charges with charge $Q=5\ \mu C$

When the first charge is at $x=10\ cm$ on the x-axis and another at x=0 (origin)

there is only Potential Energy

Total Energy(E)=Kinetic Energy(K)+Potential Energy(U)

$U_i=\frac{kQ\cdot Q}{r}$

$U_i=\frac{9\times 10^9\times (5\times 10^{-6})^2}{0.1}$

$U_i=2.25\ J$

$E=2.25$

When First charge is at $x=90$

There will be both Kinetic energy and potential Energy

$U_f=\frac{kQ\cdot Q}{r}$

$U_f=\frac{9\times 10^9\times (5\times 10^{-6})^2}{0.9}$

$U_f=0.25\ J$

As total Energy is constant therefore Kinetic Energy is

$k=E-U_f$

$k=2.25-0.25=2\ J$