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V A B = V B V A = R B E · d l + R A E · d l

which simplifies to

V B V A = A B E · d 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] .

The figure shows a charge q equidistant from two points, A and B.
The arc for calculating the potential difference between two points that are equidistant from a point charge at the origin.

To do this, we integrate around an arc of the circle of constant radius r between A and B , which means we let d l = r φ ^ d φ , while using E = k q r 2 r ^ . Thus,

Δ V A B = V B V A = A B E · d l

for this system becomes

V B V A = A B k q r 2 r ^ · r φ ^ d φ .

However, r ^ · φ ^ = 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 × 10 6 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. We can use the equation V A B = E d to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

V A B = E d .

Entering the given values for E and d gives

V A B = ( 3.0 × 10 6 V/m ) ( 0.025 m ) = 7.5 × 10 4 V

or

V A B = 75 kV .

(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] ).

The first photo shows a spark chamber and the second photo shows its operation.
A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). This form of detector is now archaic and no longer in use except for demonstration purposes. (credit b: modification of work by Jack Collins)
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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 - μ 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 = V A B d . Once we know the electric field strength, we can find the force on a charge by using F = q E . Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, F = q E .

Solution

  1. The expression for the magnitude of the electric field between two uniform metal plates is
    E = V A B 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 A B and the plate separation of 0.0400 m, we obtain
    E = 25.0 kV 0.0400 m = 6.25 × 10 5 V/m .
  2. The magnitude of the force on a charge in an electric field is obtained from the equation
    F = q E .

    Substituting known values gives
    F = ( 0.500 × 10 −6 C ) ( 6.25 × 10 5 V/m ) = 0.313 N .

Significance

Note that the units are newtons, since 1 V/m = 1 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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Questions & Answers

What is differential form of Gauss's law?
Rohit Reply
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
OLUWA Reply
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
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
Aditya
determine absolute zero
OFERE Reply
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
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
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
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
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
Ifeoma Reply
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.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
Practice Key Terms 4

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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