Conservation of Energy
Conservation of Energy
The total energy of a system is conserved if there is no net addition (or subtraction) of work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.
Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, \(\text{KE}+\text{PE = constant}\). A loss of PE of a charged particle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as
\(\text{KE}+\text{PE = constant}\)
or
\({\text{KE}}_{i}+{\text{PE}}_{i}{\text{= KE}}_{f}+{\text{PE}}_{f},\)
where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.
Example: Electrical Potential Energy Converted to Kinetic Energy
Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)
Strategy
We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be \({\text{KE}}_{i}=0,\phantom{\rule{0.25em}{0ex}}{\text{KE}}_{f}=\frac{1}{2}{\mathrm{mv}}^{2},\phantom{\rule{0.25em}{0ex}}{\text{PE}}_{i}=\mathrm{qV}{\text{, and PE}}_{f}=0.\)
Solution
Conservation of energy states that
\({\text{KE}}_{i}+{\text{PE}}_{i}{\text{= KE}}_{f}+{\text{PE}}_{f}\text{.}\)
Entering the forms identified above, we obtain
\(\text{qV}=\cfrac{{\text{mv}}^{2}}{\text{2}}\text{.}\)
We solve this for \(v\):
\(v=\sqrt{\cfrac{2\text{qV}}{m}}\text{.}\)
Entering values for \(q,\phantom{\rule{0.25em}{0ex}}V\text{, and}\phantom{\rule{0.25em}{0ex}}m\) gives
\(\begin{array}{lll}v& =& \sqrt{\cfrac{2(–1.60×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{C})(\text{–100 J/C})}{9.11×{\text{10}}^{\text{–31}}\phantom{\rule{0.25em}{0ex}}\text{kg}}}\\ & =& 5.93×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s.}\end{array}\)
Discussion
Note that both the charge and the initial voltage are negative, as in this figure. From the discussions in Electric Charge and Electric Field, we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Those higher voltages produce electron speeds so great that relativistic effects must be taken into account. That is why a low voltage is considered (accurately) in this example.
This lesson is part of:
Electric Potential and Electric Field