# 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,

What is differential form of Gauss's law?
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
kya ye b.sc ka hai agar haa to konsa part
what is charge quantization
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
determine absolute zero
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
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this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
what is the weight of the earth in space
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
explain the lack of symmetry in the field of the parallel capacitor
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe