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  • 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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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