# 7.2 Electric potential and potential difference  (Page 6/12)

 Page 6 / 12
${V}_{AB}={V}_{B}-{V}_{A}=\text{−}{\int }_{R}^{B}\stackrel{\to }{\text{E}}·d\stackrel{\to }{\text{l}}+{\int }_{R}^{A}\stackrel{\to }{\text{E}}·d\stackrel{\to }{\text{l}}$

which simplifies to

${V}_{B}-{V}_{A}=\text{−}{\int }_{A}^{B}\stackrel{\to }{\text{E}}·d\stackrel{\to }{\text{l}}.$

As a demonstration, from this we may calculate the potential difference between two points ( A and B ) equidistant from a point charge q at the origin, as shown in [link] .

To do this, we integrate around an arc of the circle of constant radius r between A and B , which means we let $d\stackrel{\to }{\text{l}}=r\stackrel{^}{\phi }d\phi ,$ while using $\stackrel{\to }{\text{E}}=\frac{kq}{{r}^{2}}\stackrel{^}{\text{r}}$ . Thus,

$\text{Δ}{V}_{AB}={V}_{B}-{V}_{A}=\text{−}{\int }_{A}^{B}\stackrel{\to }{\text{E}}·d\stackrel{\to }{\text{l}}$

for this system becomes

${V}_{B}-{V}_{A}=\text{−}{\int }_{A}^{B}\frac{kq}{{r}^{2}}\stackrel{^}{\text{r}}·r\stackrel{^}{\phi }d\phi .$

However, $\stackrel{^}{\text{r}}·\stackrel{^}{\phi }=0$ and therefore

${V}_{B}-{V}_{A}=0.$

This result, that there is no difference in potential along a constant radius from a point charge, will come in handy when we map potentials.

## What is the highest voltage possible between two plates?

Dry air can support a maximum electric field strength of about $3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{V/m}\text{.}$ Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?

## Strategy

We are given the maximum electric field E between the plates and the distance d between them. We can use the equation ${V}_{AB}=Ed$ to calculate the maximum voltage.

## Solution

The potential difference or voltage between the plates is

${V}_{AB}=Ed.$

Entering the given values for E and d gives

${V}_{AB}=\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{V/m}\right)\left(0.025\phantom{\rule{0.2em}{0ex}}\text{m}\right)=7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{V}$

or

${V}_{AB}=75\phantom{\rule{0.2em}{0ex}}\text{kV}\text{.}$

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

## Significance

One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5-cm (1-in.) gap, or 150 kV for a 5-cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage can cause a spark if there are spines on the surface, since sharp points have larger field strengths than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up with static electricity on dry days ( [link] ).

## Field and force inside an electron gun

An electron gun ( [link] ) has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. (a) What is the electric field strength between the plates? (b) What force would this field exert on a piece of plastic with a $0.500\text{-}\mu \text{C}$ charge that gets between the plates?

## Strategy

Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression $E=\frac{{V}_{AB}}{d}$ . Once we know the electric field strength, we can find the force on a charge by using $\stackrel{\to }{\text{F}}=q\stackrel{\to }{\text{E}}.$ Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, $F=qE$ .

## Solution

1. The expression for the magnitude of the electric field between two uniform metal plates is
$E=\frac{{V}_{AB}}{d}.$

Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for ${V}_{AB}$ and the plate separation of 0.0400 m, we obtain
$E=\frac{25.0\phantom{\rule{0.2em}{0ex}}\text{kV}}{0.0400\phantom{\rule{0.2em}{0ex}}\text{m}}=6.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{V/m}\text{.}$
2. The magnitude of the force on a charge in an electric field is obtained from the equation
$F=qE.$

Substituting known values gives
$F=\left(0.500\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(6.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{V/m}\right)=0.313\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$

## Significance

Note that the units are newtons, since $1\phantom{\rule{0.2em}{0ex}}\text{V/m}=1\phantom{\rule{0.2em}{0ex}}\text{N/C}$ . Because the electric field is uniform between the plates, the force on the charge is the same no matter where the charge is located between the plates.

A closely wound search coil has an area of 4cm^2,1000 turns and a resistance of 40ohm. It is connected to a ballistic galvanometer whose resistance is 24 ohm. When coil is rotated from a position parallel to uniform magnetic field to one perpendicular to field,the galvanometer indicates a charge
Using Kirchhoff's rules, when choosing your loops, can you choose a loop that doesn't have a voltage?
how was the check your understand 12.7 solved?
Who is ISSAAC NEWTON
he's the father of 3 newton law
Hawi
he is Chris Issaac's father :)
Ethem
how to name covalent bond
Who is ALEXANDER BELL
LOAK
what do you understand by the drift voltage
what do you understand by drift velocity
Brunelle
nothing
Gamal
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.
graviton
drift velocity
graviton
what is an electromotive force?
It is the amount of other forms of energy converted into electrical energy per unit charge that flow through it.
Brunelle
How electromotive force is differentiated from the terminal voltage?
Danilo
in the emf power is generated while in the terminal pd power is lost.
Brunelle
what is then chemical name of NaCl
sodium chloride
Azam
sodium chloride
Brunelle
Sodium Chloride.
Ezeanyim
How can we differentiate between static point and test charge?
Wat is coplanar in physics
two point charges +30c and +10c are separated by a distance of 80cm,compute the electric intensity and force on a +5×10^-6c charge place midway between the charges
0.0844kg
Humble
what is the difference between temperature and heat
Heat is the condition or quality of being hot While Temperature is ameasure of cold or heat, often measurable with a thermometer
Abdul
Temperature is the one of heat indicators of materials that can be measured with thermometers, and Heat is the quantity of calor content in material that can be measured with calorimetry.
Gamma
the average kinetic energy of molecules is called temperature. heat is the method or mode to transfer energy to molecules of an object but randomly, while work is the method to transfer energy to molecules in such manner that every molecules get moved in one direction.
2. A brass rod of length 50cm and diameter 3mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°c( degree Celsius) if the original length are 40°c(degree Celsius) is there at thermal stress developed at the junction? The end of the rod are free to expand (coefficient of linear expansion of brass = 2.0×10^-5, steel=1.2×10^-5k^1)
A charge insulator can be discharged by passing it just above a flame. Explain.
of the three vectors in the equation F=qv×b which pairs are always at right angles?