Given a point charge
$q=+2.0\phantom{\rule{0.2em}{0ex}}\text{nC}$ at the origin, calculate the potential difference between point
${P}_{1}$ a distance
$a=4.0\phantom{\rule{0.2em}{0ex}}\text{cm}$ from
q , and
${P}_{2}$ a distance
$b=12.0\phantom{\rule{0.2em}{0ex}}\text{cm}$ from
q , where the two points have an angle of
$\phi =24\text{\xb0}$ between them (
[link] ).
Strategy
Do this in two steps. The first step is to use
${V}_{B}-{V}_{A}=\text{\u2212}{\displaystyle {\int}_{A}^{B}\overrightarrow{\text{E}}\xb7d\overrightarrow{\text{l}}}$ and let
$A=a=4.0\phantom{\rule{0.2em}{0ex}}\text{cm}$ and
$B=b=12.0\phantom{\rule{0.2em}{0ex}}\text{cm},$ with
$d\overrightarrow{\text{l}}=d\overrightarrow{\text{r}}=\widehat{\text{r}}dr$ and
$\overrightarrow{\text{E}}=\frac{kq}{{r}^{2}}\widehat{\text{r}}.$ Then perform the integral. The second step is to integrate
${V}_{B}-{V}_{A}=\text{\u2212}{\displaystyle {\int}_{A}^{B}\overrightarrow{\text{E}}\xb7d\overrightarrow{\text{l}}}$ around an arc of constant radius
r , which means we let
$d\overrightarrow{\text{l}}=r\widehat{\phi}d\phi $ with limits
$0\le \phi \le 24\text{\xb0},$ still using
$\overrightarrow{\text{E}}=\frac{kq}{{r}^{2}}\widehat{\text{r}}$ . Then add the two results together.
Solution
For the first part,
${V}_{B}-{V}_{A}=\text{\u2212}{\displaystyle {\int}_{A}^{B}\overrightarrow{\text{E}}\xb7d\overrightarrow{\text{l}}}$ for this system becomes
${V}_{b}-{V}_{a}=\text{\u2212}{\displaystyle {\int}_{a}^{b}\frac{kq}{{r}^{2}}\widehat{\text{r}}\xb7\widehat{\text{r}}dr}$ which computes to
For the second step,
${V}_{B}-{V}_{A}=\text{\u2212}{\displaystyle {\int}_{A}^{B}\overrightarrow{\text{E}}\xb7d\overrightarrow{\text{l}}}$ becomes
$\text{\Delta}V=\text{\u2212}{\displaystyle {\int}_{0}^{24\text{\xb0}}\frac{kq}{{r}^{2}}\widehat{\text{r}}\xb7r\widehat{\phi}d\phi}$ , but
$\widehat{\text{r}}\xb7\widehat{\phi}=0$ and therefore
$\text{\Delta}V=0.$ Adding the two parts together, we get 300 V.
Significance
We have demonstrated the use of the integral form of the potential difference to obtain a numerical result. Notice that, in this particular system, we could have also used the formula for the potential due to a point charge at the two points and simply taken the difference.
Check Your Understanding From the examples, how does the energy of a lightning strike vary with the height of the clouds from the ground? Consider the cloud-ground system to be two parallel plates.
Given a fixed maximum electric field strength, the potential at which a strike occurs increases with increasing height above the ground. Hence, each electron will carry more energy. Determining if there is an effect on the total number of electrons lies in the future.
Before presenting problems involving electrostatics, we suggest a problem-solving strategy to follow for this topic.
Problem-solving strategy: electrostatics
Examine the situation to determine if static electricity is involved; this may concern separated stationary charges, the forces among them, and the electric fields they create.
Identify the system of interest. This includes noting the number, locations, and types of charges involved.
Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. Determine whether the Coulomb force is to be considered directly—if so, it may be useful to draw a free-body diagram, using electric field lines.
Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is important to distinguish the Coulomb force
F from the electric field
E , for example.
Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines as requested.
Examine the answer to see if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?
Summary
Electric potential is potential energy per unit charge.
The potential difference between points
A and
B ,
${V}_{B}-{V}_{A},$ that is, the change in potential of a charge
q moved from
A to
B , is equal to the change in potential energy divided by the charge.
Potential difference is commonly called voltage, represented by the symbol
$\text{\Delta}V$ :
$\text{\Delta}V=\frac{\text{\Delta}U}{q}\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}\text{\Delta}U=q\text{\Delta}V.$
An electron-volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,
$1\phantom{\rule{0.2em}{0ex}}\text{eV}=\left(1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(1\phantom{\rule{0.2em}{0ex}}\text{V}\right)=\left(1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{\phantom{\rule{0.2em}{0ex}}10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(1\phantom{\rule{0.2em}{0ex}}\text{J/C}\right)=1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
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
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
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
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.