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

Step-by-step 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=(kQ\cdot Q)/(r)`

`U_i=(9* 10^9* (5* 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=(kQ\cdot Q)/(r)`

`U_f=(9* 10^9* (5* 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`