# 19.2 Electric potential in a uniform electric field

 Page 1 / 5
• Describe the relationship between voltage and electric field.
• Derive an expression for the electric potential and electric field.
• Calculate electric field strength given distance and voltage.

In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field $\mathbf{\text{E}}$ is produced by placing a potential difference (or voltage) $\Delta V$ across two parallel metal plates, labeled A and B. (See [link] .) Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either $\Delta V$ or $\mathbf{\text{E}}$ can be used to describe any charge distribution. $\Delta V$ is most closely tied to energy, whereas $\mathbf{\text{E}}$ is most closely related to force. $\Delta V$ is a scalar    quantity and has no direction, while $\mathbf{\text{E}}$ is a vector    quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by $E$ below.) The relationship between $\Delta V$ and $\mathbf{\text{E}}$ is revealed by calculating the work done by the force in moving a charge from point A to point B. But, as noted in Electric Potential Energy: Potential Difference , this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in [link] to move a positive charge $q$ from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

$W=\phantom{\rule{0.25em}{0ex}}–\Delta \text{PE}=\phantom{\rule{0.25em}{0ex}}–q\Delta V.$

The potential difference between points A and B is

$–\Delta V=\phantom{\rule{0.25em}{0ex}}–\left({V}_{\text{B}}–{V}_{\text{A}}\right)={V}_{\text{A}}–{V}_{\text{B}}={V}_{\text{AB}}.$

Entering this into the expression for work yields

$W={\text{qV}}_{\text{AB}}.$

Work is $W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ ; here $\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =1$ , since the path is parallel to the field, and so $W=\text{Fd}$ . Since $F=\text{qE}$ , we see that $W=\text{qEd}$ . Substituting this expression for work into the previous equation gives

$\mathrm{qEd}={\text{qV}}_{\text{AB}}.$

The charge cancels, and so the voltage between points A and B is seen to be

$\begin{array}{c}\left(\begin{array}{c}{V}_{\text{AB}}=\mathrm{Ed}\\ E=\frac{{V}_{\text{AB}}}{d}\end{array}}\text{(uniform}\phantom{\rule{0.25em}{0ex}}E\phantom{\rule{0.25em}{0ex}}\text{- field only),}\end{array}$

where $d$ is the distance from A to B, or the distance between the plates in [link] . Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid:

$\text{1 N}/C=\text{1 V}/m.$

## Voltage between points a and b

$\begin{array}{c}\left(\begin{array}{c}{V}_{\text{AB}}=\mathrm{Ed}\\ E=\frac{{V}_{\text{AB}}}{d}\end{array}}\text{(uniform}\phantom{\rule{0.25em}{0ex}}E\phantom{\rule{0.25em}{0ex}}\text{- field only),}\end{array}$

where $d$ is the distance from A to B, or the distance between the plates.

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

Dry air will support a maximum electric field strength of about $3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}$ . 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. The equation ${V}_{\text{AB}}=\mathrm{Ed}$ can thus be used to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

${\text{V}}_{\text{AB}}=\mathrm{Ed}.$

Entering the given values for $E$ and $d$ gives

${V}_{\text{AB}}=\left(3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}\right)\left(0.025 m\right)=7.5×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}V$

or

${V}_{\text{AB}}=\text{75 kV}.$

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

Discussion

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 will cause a spark if there are points on the surface, since points create greater fields 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, say with static electricity, on dry days.

In Inelastic collision cunculate the vilocity
elucidate
Usman
explain how a body becomes electrically charged based on the presence of charged particles
induction
babar
induction
DEMGUE
definitely by induction
Raymond
induction
Raymond
induction
Shah
induction
Korodhso
please why does a needle sinks in water
DEMGUE
induction
Korodhso
induction
Auwal
what are the calculations of Newton's third law of motiow
what is dark matter
(in some cosmological theories) non-luminous material which is postulated to exist in space and which could take either of two forms: weakly interacting particles ( cold dark matter ) or high-energy randomly moving particles created soon after the Big Bang ( hot dark matter ).
Usman
if the mass of a trolley is 0.1kg. calculate the weight of plasticine that is needed to compensate friction. (take g=10m/s and u=0.2)
what is a galaxy
a galaxy is a type of phone e.g samsung galaxy there are diff types of samsung galaxy there is s5 s6 s7 s8 s9
lasisi
what isflow rate of volume
flow rate is the volume of fluid which passes per unit time;
Rev
flow rate or discharge represnts the flow passing in unit volume per unit time
bhat
When two charges q1 and q2 are 6 and 5 coulomb what is ratio of force
incomplete question
lasisi
When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?
nehemiah
why is it proportional
i don't know
y
nehemiah
what are the relationship between distance and displacement
They are interchangeable.
Shii
Distance is scalar, displacement is vector because it must involve a direction as well as a magnitude. distance is the measurement of where you are and where you were displacement is a measurement of the change in position
Shii
Thanks a lot
Usman
I'm beginner in physics so I can't reason why v=u+at change to v2=u2+2as and vice versa
Usman
what is kinematics
praveen
kinematics is study of motion without considering the causes of the motion
Theo
The study of motion without considering the cause 0f it
Usman
why electrons close to the nucleus have less energy and why do electrons far from the nucleus have more energy
Theo
thank you frds
praveen
plz what is the third law of thermodynamics
third law of thermodynamics states that at 0k the particles will collalse its also known as death of universe it was framed at that time when it waa nt posible to reach 0k but it was proved wrong
bhat
I have not try that experiment but I think it will magnet....
Hey Rev. it will
Jeff
I do think so, it will
Chidera
yes it will
lasisi
If a magnet is in a pool of water, would it be able to have a magnetic field?.
yes Stella it would
Jeff
formula for electric current
Fokoua
what are you given?
Kudzy
what is current
Fokoua
I=q/t
saifullahi
Current is the flow of electric charge per unit time.
saifullahi
What are semi conductors
saifullahi
materials that allows charge to flow at varying conditions, temperature for instance.
Mokua
these are materials which have electrical conductivity greater than the insulators but less than metal, in these materials energy band Gap is very narrow as compared to insulators
Sunil
materials that allows charge to flow at varying conditions, temperature for instance.
Obasi
wao so awesome
Fokoua
At what point in the oscillation of beam will a body leave it?
Atambiri
what is gravitational force