Question

A particle (charge 7.5 µC) is released from rest at a point on the x axis, x = 10 cm. It begins to move due to the presence of a 2.0-µC charge which remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point x = 1.0 m?

Answer 1

Answer:

1.215J

Explanation:

Using the law of energy conservation, the potential energy is equal to the kinetic energy of the particle during motion:

$U_1=_qV$

#the V of the 2.0-µC is expressed as:

$V=\frac{1}{4\pi \epsilon_o r}$

We substitute the given values from the question:

##When the moving charge reaches 0.1meter the energy becomes

$U_1=7.5\times 10^{-6}\times \frac{2.0\times10^{-6}}{4\pi 8.85\times 10^{-12}\times 0.1\\}\\=1.35J$

#When the moving charge reaches 1.0 meter the energy becomes

$U_2=7.5\times 10^{-6}\times \frac{2.0\times10^{-6}}{4\pi 8.85\times 10^{-12}\times 1.0}\\\\=0.135J$

Hence, the change in energy is $U_1-U_2=1.35-0.135=1.215J$