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?(A) 3.0 J(B) 1.8 J(C) 2.4 J(D) 1.2 J(E) 1.4 J

Answer 1

The correct answer is option 'D': 1.2 Joules

Step-by-step explanation:

Since the energy associated with the electric field is conservative thus we conclude that the

Kinetic energy of the moving particle equals the change in electrostatic potential energy.

For the initial case

`U_1=qV`

Now V associated with the 2.0μC charge is

`V=\frac{}{4\pi \epsilon _o\cdot r}`

Applying the given values we get

`U_1=7.5* 10^(-6)* (2.0* 10^(-6))/(4\pi * 8.85* 10^(-12)* 0.1)\\\\\therefore U_1=1.35J`

When the moving charge reaches x = 1.0 meter the energy becomes

`U_2=7.5* 10^(-6)* (2.0* 10^(-6))/(4\pi * 8.85* 10^(-12)* 1)\\\\\therefore U_2=0.135J`

Thus the change in energy is
`U_1-U_2=1.35-0.135=1.2J`