# 7.2 Electric potential and potential difference  (Page 4/12)

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## Conservation of energy

The total energy of a system is conserved if there is no net addition (or subtraction) due to 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, $K+U=\text{constant}\text{.}$ A loss of U for a charged particle becomes an increase in its K . Conservation of energy is stated in equation form as

$K+U=\text{constant}$

or

${K}_{\text{i}}+{U}_{\text{i}}={K}_{\text{f}}+{U}_{\text{f}}$

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.

## Electrical potential energy converted into 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 ${K}_{\text{i}}=0,{K}_{\text{f}}=\frac{1}{2}m{v}^{2},{U}_{\text{i}}=qV,{U}_{\text{f}}=0.$

## Solution

Conservation of energy states that

${K}_{\text{i}}+{U}_{\text{i}}={K}_{\text{f}}+{U}_{\text{f}}.$

Entering the forms identified above, we obtain

$qV=\frac{m{v}^{2}}{2}.$

We solve this for v :

$v=\sqrt{\frac{2qV}{m}}.$

Entering values for q , V , and m gives

$v=\sqrt{\frac{2\left(-1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(-100\phantom{\rule{0.2em}{0ex}}\text{J/C}\right)}{9.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\phantom{\rule{0.2em}{0ex}}\text{kg}}}=5.93\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m/s}\text{.}$

## Significance

Note that both the charge and the initial voltage are negative, as in [link] . From the discussion of 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. These higher voltages produce electron speeds so great that effects from special relativity must be taken into account and hence are reserved for a later chapter ( Relativity ). That is why we consider a low voltage (accurately) in this example.

Check Your Understanding How would this example change with a positron? A positron is identical to an electron except the charge is positive.

It would be going in the opposite direction, with no effect on the calculations as presented.

## Voltage and electric field

So far, we have explored the relationship between voltage and energy. Now we want to explore the relationship between voltage and electric field. We will start with the general case for a non-uniform $\stackrel{\to }{\text{E}}$ field. Recall that our general formula for the potential energy of a test charge q at point P relative to reference point R is

${U}_{P}=\text{−}{\int }_{R}^{P}\stackrel{\to }{\text{F}}·d\stackrel{\to }{\text{l}}.$

When we substitute in the definition of electric field $\left(\stackrel{\to }{\text{E}}=\stackrel{\to }{\text{F}}\text{/}q\right),$ this becomes

${U}_{P}=\text{−}q{\int }_{R}^{P}\stackrel{\to }{\text{E}}·d\stackrel{\to }{\text{l}}.$

Applying our definition of potential $\left(V=U\text{/}q\right)$ to this potential energy, we find that, in general,

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