<< Chapter < Page Chapter >> Page >
  • Explain equipotential lines and equipotential surfaces.
  • Describe the action of grounding an electrical appliance.
  • Compare electric field and equipotential lines.

We can represent electric potentials (voltages) pictorially, just as we drew pictures to illustrate electric fields. Of course, the two are related. Consider [link] , which shows an isolated positive point charge and its electric field lines. Electric field lines radiate out from a positive charge and terminate on negative charges. While we use blue arrows to represent the magnitude and direction of the electric field, we use green lines to represent places where the electric potential is constant. These are called equipotential lines in two dimensions, or equipotential surfaces in three dimensions. The term equipotential is also used as a noun, referring to an equipotential line or surface. The potential for a point charge is the same anywhere on an imaginary sphere of radius r size 12{r} {} surrounding the charge. This is true since the potential for a point charge is given by V = kQ / r size 12{V= ital "kQ"/r} {} and, thus, has the same value at any point that is a given distance r size 12{r} {} from the charge. An equipotential sphere is a circle in the two-dimensional view of [link] . Since the electric field lines point radially away from the charge, they are perpendicular to the equipotential lines.

The figure shows a positive charge Q at the center of four concentric circles of increasing radii. The electric potential is the same along each of the circles, called equipotential lines. Straight lines representing electric field lines are drawn from the positive charge to intersect the circles at various points. The equipotential lines are perpendicular to the electric field lines.
An isolated point charge Q size 12{Q} {} with its electric field lines in blue and equipotential lines in green. The potential is the same along each equipotential line, meaning that no work is required to move a charge anywhere along one of those lines. Work is needed to move a charge from one equipotential line to another. Equipotential lines are perpendicular to electric field lines in every case.

It is important to note that equipotential lines are always perpendicular to electric field lines . No work is required to move a charge along an equipotential, since Δ V = 0 size 12{?`V`=`0} {} . Thus the work is

W = –Δ PE = q Δ V = 0 . size 12{W=-?"PE"=-q?V=0} {}

Work is zero if force is perpendicular to motion. Force is in the same direction as E size 12{E} {} , so that motion along an equipotential must be perpendicular to E size 12{E} {} . More precisely, work is related to the electric field by

W = Fd cos θ = qEd cos θ = 0 . size 12{W=`` ital "Fd""cos"?`=` ital "qEd""cos"?`=0} {}

Note that in the above equation, E size 12{E} {} and F size 12{F} {} symbolize the magnitudes of the electric field strength and force, respectively. Neither q size 12{q} {} nor E nor d is zero, and so cos θ must be 0, meaning θ size 12{?} {} must be 90º . In other words, motion along an equipotential is perpendicular to E size 12{E} {} .

One of the rules for static electric fields and conductors is that the electric field must be perpendicular to the surface of any conductor. This implies that a conductor is an equipotential surface in static situations . There can be no voltage difference across the surface of a conductor, or charges will flow. One of the uses of this fact is that a conductor can be fixed at zero volts by connecting it to the earth with a good conductor—a process called grounding    . Grounding can be a useful safety tool. For example, grounding the metal case of an electrical appliance ensures that it is at zero volts relative to the earth.

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
Syamthanda Reply
hey , can you please explain oxidation reaction & redox ?
Boitumelo Reply
hey , can you please explain oxidation reaction and redox ?
Boitumelo
for grade 12 or grade 11?
Sibulele
the value of V1 and V2
Tumelo Reply
advantages of electrons in a circuit
Rethabile Reply
we're do you find electromagnetism past papers
Ntombifuthi
what a normal force
Tholulwazi Reply
it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface
Sihle
what is physics?
Petrus Reply
what is the half reaction of Potassium and chlorine
Anna Reply
how to calculate coefficient of static friction
Lisa Reply
how to calculate static friction
Lisa
How to calculate a current
Tumelo
how to calculate the magnitude of horizontal component of the applied force
Mogano
How to calculate force
Monambi
a structure of a thermocouple used to measure inner temperature
Anna Reply
a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4
Amahle Reply
How is energy being used in bonding?
Raymond Reply
what is acceleration
Syamthanda Reply
a rate of change in velocity of an object whith respect to time
Khuthadzo
how can we find the moment of torque of a circular object
Kidist
Acceleration is a rate of change in velocity.
Justice
t =r×f
Khuthadzo
how to calculate tension by substitution
Precious Reply
hi
Shongi
hi
Leago
use fnet method. how many obects are being calculated ?
Khuthadzo
khuthadzo hii
Hulisani
how to calculate acceleration and tension force
Lungile Reply
you use Fnet equals ma , newtoms second law formula
Masego
please help me with vectors in two dimensions
Mulaudzi Reply
how to calculate normal force
Mulaudzi
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask