# 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.

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