Chapter 33, Problem 39A

a. If you do 12 J of work to push 0.001 C of charge from point A to point B in an electric field, what is the voltage difference between points A and B?

b. When the charge is released, what will be its kinetic energy as it flies back past its starting point A? What principle guides your answer?

To determine

To explain: The voltage difference between points $\text{A}$ and $\text{B}$

Explanation of Solution

Work charge particle (q = 0.001 C) moved between points A and B is 12 J .

The voltage difference between A and B,

  $\begin{array}{c}V=\frac{U}{q}\\ =\frac{\left(12\text{ J}\right)}{\left(0.001\text{ C}\right)}\\ =12000\text{ J/C }\\ =12000\text{ V}\end{array}$

To determine

To find: The kinetic energy of charged particle and the principle behind it.

Explanation of Solution

Principle is conservation of energy

From the conservation of energy. Total energy of the system remain constant. So decrease in potential energy would increase the kinetic energy. The electric potential energy 12 J will be converted to 12 J of kinetic energy.