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,
$1\phantom{\rule{0.2em}{0ex}}\text{eV}=\left(1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(1\phantom{\rule{0.2em}{0ex}}\text{V}\right)=\left(1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{\phantom{\rule{0.2em}{0ex}}10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(1\phantom{\rule{0.2em}{0ex}}\text{J/C}\right)=1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
Questions & Answers
A mass 'm' is attached to a spring oscillates every 5 second. If the mass is increased by a 5 kg, the period increases by 3 second. Find its initial mass 'm'
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well when you apply a small electric field to a conductor that causes to add a little velocity to charged particle than usual, which become their average speed, that is what we call a drift.
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