A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases potential energy is converted to another form. Work is done by a force, but since this force is conservative, we can write W = –ΔPE.When a free positive charge q is accelerated by an electric field, such as shown in Figure 1, it is given kinetic energy.
The process is analogous to an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. Let us explore the work done on a charge q by the electric field in this process, so that we may develop a definition of electric potential energy.The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is usually easier to deal with the potential energy (because it depends only on position) than to calculate the work directly.We use the letters PE to denote electric potential energy, which has units of joules (J). The change in potential energy, ΔPE, is crucial, since the work done by a conservative force is the negative of the change in potential energy; that is, W = –ΔPE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative ΔPE.
Electric Potential At A Point
There must be a minus sign in front of ΔPE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.
Making Connections: Energy UnitsThe electron volt (eV) is the most common energy unit for submicroscopic processes. This will be particularly noticeable in the chapters on modern physics. Energy is so important to so many subjects that there is a tendency to define a special energy unit for each major topic. There are, for example, calories for food energy, kilowatt-hours for electrical energy, and therms for natural gas energy.The electron volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it is given an energy of 30 keV (30,000 eV) and it can break up as many as 6000 of these molecules (30,000 eV ÷ 5 eV per molecule= 6000 molecules). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can, thus, produce significant biological damage.
Electrical Potential Versus Potential Energy
Conservation of EnergyThe total energy of a system is conserved if there is no net addition (or subtraction) of work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, KE+PE = constant. A loss of PE of a charged particle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as KE + PE = constant or KE i + PE i = KE f + PE f, where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.
Conceptual Questions. Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?. If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force?
Explain. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?. Voltages are always measured between two points.
Why?. How are units of volts and electron volts related? How do they differ? Problems & Exercises.
Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 1.67 × 10 −27 kg. An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce x rays. Non-relativistically, what would be the maximum speed of these electrons?.
A bare helium nucleus has two positive charges and a mass of 6.64 × 10 −27 kg. (a) Calculate its kinetic energy in joules at 2.00% of the speed of light. (b) What is this in electron volts? (c) What voltage would be needed to obtain this energy?. Integrated Concepts. Singly charged gas ions are accelerated from rest through a voltage of 13.0 V. At what temperature will the average kinetic energy of gas molecules be the same as that given these ions?.
Integrated Concepts. The temperature near the center of the Sun is thought to be 15 million degrees Celsius (1.5 × 10 7 ºC). Through what voltage must a singly charged ion be accelerated to have the same energy as the average kinetic energy of ions at this temperature?. Integrated Concepts. (a) What is the average power output of a heart defibrillator that dissipates 400 J of energy in 10.0 ms?
(b) Considering the high-power output, why doesn’t the defibrillator produce serious burns?. Integrated Concepts. A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00 × 10 2 MV. (a) What energy was dissipated? (b) What mass of water could be raised from 15ºC to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.
Integrated Concepts. A 12.0 V battery-operated bottle warmer heats 50.0 g of glass, 2.50 × 10 2 g of baby formula, and 2.00 × 10 2 g of aluminum from 20.0ºC to 90.0ºC. (a) How much charge is moved by the battery? (b) How many electrons per second flow if it takes 5.00 min to warm the formula? (Hint: Assume that the specific heat of baby formula is about the same as the specific heat of water.). Integrated Concepts. A battery-operated car utilizes a 12.0 V system.
Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00 × 10 2 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00 × 10 2 N force for an hour. Integrated Concepts. Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. (a) Calculate the potential energy of two singly charged nuclei separated by 1.00 × 10 −12 m by finding the voltage of one at that distance and multiplying by the charge of the other. (b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?. Unreasonable Results. (a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it.
(b) What is unreasonable about this result? (c) Which assumptions are responsible?. Construct Your Own Problem. Consider a battery used to supply energy to a cellular phone. Construct a problem in which you determine the energy that must be supplied by the battery, and then calculate the amount of charge it must be able to move in order to supply this energy.
Among the things to be considered are the energy needs and battery voltage. You may need to look ahead to interpret manufacturer’s battery ratings in ampere-hours as energy in joules.Glossaryelectric potential: potential energy per unit chargepotential difference (or voltage): change in potential energy of a charge moved from one point to another, divided by the charge; units of potential difference are joules per coulomb, known as voltelectron volt: the energy given to a fundamental charge accelerated through a potential difference of one voltmechanical energy: sum of the kinetic energy and potential energy of a system; this sum is a constant. Selected Solutions to Problems & Exercises1. 42.84. 1. Dawn of war dark crusade cheat engine. 00 × 10 5 K6. (a) 4 × 10 4 W; (b) A defibrillator does not cause serious burns because the skin conducts electricity well at high voltages, like those used in defibrillators. The gel used aids in the transfer of energy to the body, and the skin doesn’t absorb the energy, but rather lets it pass through to the heart.8. (a) 7.40 × 10 3 C; (b) 1.54 × 10 20 electrons per second9. 3.89 × 10 6 C11. (a) 1.44 × 10 12 V; (b) This voltage is very high.
A 10.0 cm diameter sphere could never maintain this voltage; it would discharge; (c) An 8.00 C charge is more charge than can reasonably be accumulated on a sphere of that size.
The problem statement, all variables and given/known dataWhat is the difference between an electric field and electric potential?2. Relevant equationsElectric Field = E = Kq/r 2Electric Potential = V = Kq/r3. The attempt at a solutionFrom the equations, I can see that the electric field strength decreases as an inverse square of r (distance) whereas the electric potential decreases linearly with distance.The electric field is a vector while the electric potential is a scalar.An electric field is the sum of any electric forces acting on a charge or group of charges. An electric field can cause a particle (proton, electron) to accelerate, as when passing through a charged capacitor in a CRT display.
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Flux is the amount of, or flow of electric field passing through an object.Both electric field and electric potential can be obtained by superposition (adding up individual field contributions or individual potential contributions).All this is great and taken from various pages in my book. But, what does it all mean? I still don't have a clear.intuition. about the difference between electric field and electric potential.I know a battery has electric potential (voltage) and there is an electric field within it between the positive and negative sides.
What else can you add to help me realize the.physical. concept of electric field vs.
Electric potential? The problem statement, all variables and given/known dataWhat is the difference between an electric field and electric potential?2. Relevant equationsElectric Field = E = Kq/r 2Electric Potential = V = Kq/r3. The attempt at a solutionFrom the equations, I can see that the electric field strength decreases as an inverse square of r (distance) whereas the electric potential decreases linearly with distance.The electric field is a vector while the electric potential is a scalar.An electric field is the sum of any electric forces acting on a charge or group of charges. An electric field can cause a particle (proton, electron) to accelerate, as when passing through a charged capacitor in a CRT display. Flux is the amount of, or flow of electric field passing through an object.Both electric field and electric potential can be obtained by superposition (adding up individual field contributions or individual potential contributions).All this is great and taken from various pages in my book. But, what does it all mean?
I still don't have a clear.intuition. about the difference between electric field and electric potential.I know a battery has electric potential (voltage) and there is an electric field within it between the positive and negative sides. What else can you add to help me realize the.physical. concept of electric field vs.
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Electric potential? The voltage at a point represents the work per unit charge needed to place any charge q there. It is analogous to saying gravitational potential is the work per unit mass needed to place mass at a height h.
(so gravitational potential would be g.h and you multiply any mass into it to find gravitational potential energy. With electric potential, you multiply q into E.d to find the electric potential energy. Notice that E and G are both fields and that h and d are both distances)E field is a vector that represents how much force and what direction that force will be per unit charge placed in that field. This is analogous to saying g is the force per unit mass for any mass at a point inside a gravitational field of g. Remember this from mechanics: FORCE = ma.
Well, force also = qE.
I'm reading about electricity on the Sparkfun website , and I read the following:'At any point in an electric field the electric potential is the amount of electric potential energy divided by the amount of charge at that point. It takes the charge quantity out of the equation and leaves us with an idea of how much potential energy specific areas of the electric field may provide.' My question is, why do we need to take the charge quantity out of the equation? It seems like dividing one by the other would give us some sort of rate (i.e.
Just like dividing distance by time gives us a rate of speed). Is that what we're trying to achieve by performing division here? Or is it closer to dividing in order to reduce a fraction, i.e. 5/15 is equal to 1/3?
Electric Potential and Electric Potential Energy with ExamplesElectric Potential and Electric Potential EnergyWe learned that in work power energy chapter, objects have potential energy because of their positions. In this case charge in an electric field has also potential energy because of its positions. Since there is a force on the charge and it does work against to this force we can say that it must have energy for doing work.
In other words, we can say that Energy required increasing the distance between two charges to infinity or vice verse. Electric potential energy is a scalar quantity and Joule is the unit of it. We use following formula to find the magnitude of EP;Be careful!. In this formula if the charges have opposite sign then, Ep becomes negative, if they are same type of charge then, Ep becomes positive. If Ep is positive then, electric potential energy is inversely proportional to the distance d. If Ep is negative then, electric potential energy is directly proportional to the distance d.In Figure 1 and Figure 2, charges repel each other, thus external forces does work for decreasing the distance between them.
On the contrary, in Figure3, charges attract each other, distance between them is decreased by electric forces, and there is no need for other external forces.Example: System given below is composed of the charges, 10q, 8q and -5q. Fin the total electric potential energy of the system.Electric PotentialElectric potential is the electric potential energy per unit charge. It is known as voltage in general, represented by V and has unit volt (joule/C).1C charge is brought to the point A from infinity. Work done here is called potential of q at A. Electric potential is found by the given formula;V=k.q/dV is a scalar quantity.
If q is negative then V becomes negative, or if q is positive then V becomes positive.Surfaces having equal potentials are called equipotential surfaces.Potential of a Charged SpherePotential at surface is equal to the potential inside the sphere. Since there is no force acting inside the sphere, work is not done to bring the charge from surface to the inside of the sphere. As the distance from the surface of the sphere increase, potential decreases. Picture given below shows the change in the potential of the sphere inside, surface and outside. As you can see, potential is constant inside and surface of the sphere; however, it decreases with the increasing distance outside it.Potential Difference between Two PointsWork done against to the electric field to move unit charge from one point to another is called potential difference between these two points.
This difference is found by differences of potential of last point from initial point. If we take the point charge from A to B, then potential difference is found by the formula;Example: Find the potential difference between points A and B, V AB in terms of kq/r?Example: If the total electric field produced by q and q' is like in the picture given below, find the electric potential of the A.
Gravitational and Electric Potential EnergyGravitationaland Electric Potential EnergyPART A.Problem:What is meant by Electric Potential,Electric Potential Energy and Change inElectric Potential Energy?Introduction:Definition of Potential#1 (from Webster's Ideal Dictionary): Something that candevelop or become actualMany times when a coach of asport is interviewed on television, if the interview is beforethe season starts, he or she indicates that the team has 'potential'.They usually do not say that their teams are 'Weak, Slowand Chicken'! What does it mean when a coach says, 'theteam has potential? 'Gravitational Potential EnergyEarly in the course you probablydiscussed the acceleration of gravity, the force of gravity andNewton's Law of Universal Gravitation. Later in this course,you probably studied potential energy.Procedure #1:You have a text book. Hold thisbook in this room (up off the table or desk). Assuming the gravitationalpotential energy at the level of the floor of this room is zero,what is the gravitational potential energy of your book?
Expressyour answer in joules.Change in Gravitational Potential EnergyProcedure #2:Assume that the level of thefloor in this room has gravitational potential energy of zerojoules. Place your textbook on one of the chairs in the room.Calculate its gravitational potential energy in joules. Now movethe book horizontally to a similar chair elsewhere in the room.Did you do work 'on the book'? Explain your answer.(Review the definition of work, if needed.)Now move the book from the chair to a laboratory table in theroom.
With the book lying on the table, calculate the book'sgravitational potential energy. Was there a change in gravitationalpotential energy? If so, what was the change in gravitationalpotential energy? Express your answer in joules.Was work done 'on the book'in moving it from the chair to the table? Explain your answer.Gravitational Potential DifferenceIn the previous activity, youhave used the fact that the change in gravitational potentialenergy is equal to the mass of the object multiplied by the accelerationdue to the earth's gravitational field multiplied by the changein the height of the object. ( D PE = mg D h) If we were to divide both side s ofthe equation by the mass, then D PE/m =g D h. This is called the difference in gravitationalpotential or the gravitational potentialdifference.At the surface of theearth there is a gravitational potential difference of about9.8 joules/kilogram for every one meter increase in elevation.Potential difference describes how much the potential energychanges if a unit test object is moved from one point to anotherin a force field.
The object does not have to move forthe potential difference to exist. The potential difference iscreated by the source that is producing the force field, notby the test object.A) A potential exists at anypoint as a result of a field. A gravitational potential (gh)exists at any point in the space associated with a gravitationalfield. Similarly, an electrical potential (Ed) exists at anypoint in the space associated with an electric field.B) If we place 'our'charge 'q' in an electric field it will have anelectric potential energy (PE = Fd = qEd). This is analogoustogravitational potential energy (PE=Fd=mgh).C) The effect of moving a massin a gravitational field creates a change in potential energyof 'our' mass 'm' such that D PE=mg D h. When work is done in moving our charge'q' in the electric field, there is a change in the electricpotential energy such that D PE = qE D d.
If we find the change in the electricpotential energy for each unit of 'our' electric charge,then D PE/ q = E D d.D) This change in electric potentialenergy/charge is measured in units of volts ( 1 volt = 1 joule/ 1 coulomb)Notice that this quantity isalso the change (or difference) in electric potential (E D d) sometimescalled the electric potential difference.E) A battery functions by creatinga difference in electric potentialenergy between two points. Definition of Potential # 2 (fromWebster's Ideal Dictionary): the degree of electrification withreference to a standardPART B.Experiment #1 -Problem:How can electric potential differencesbe detected?Materials:Pyrex baking pan or clear plastictray, voltmeter, connecting wires with alligatorclips, two electrical conductors that sink in water, power supply,graph paper and salt.Procedure:1.
Place the graph paper on atable and center the baking dish on the grid. Hook one alligatorlead to the positive terminal and another wire to the negativeterminal of the supply.
Attach each of the wire ends to one ofthe conductors, and separate the conductors in the dish.2. Fill the dish one forth fullwith water.3. Connect the power supply tothe electrodes as illustrated and adjust thepotential to approximately 10 volts. Attach one lead of the voltmeterto one ofthe electrodes and let the other lead of the voltmeter be the'probe' that will beused to plot the equipotentials. As a preliminary check, placethe probe on the '+' electrode and not the full scalereading. Then move the probe into the salt water very near the'+' electrode. You should note only a small changeint he reading in the voltmeter.
If a large change is observed,this means either the water is not sufficiently conductive oryou should use a voltmeter with a higher internalresistance. Now move the probe between the electrodes and youshould observe afairly linear change in voltage with distance for the '-'electrode. You are nowready to plot equipotentials.4. Draw the locations of theconductors on the graph paper an d label them with the voltagereadings of your voltmeter.5. With your positive voltmeter probe in the water, note the voltage reading. Move the probein the water, keeping the voltage reading at the same value.How far does this path go? Sketch this pattern on your graphpaper and label the line with the voltage you chose.6.
Choose other voltage valuesand similarly sketch their patterns on the graph paper untilyou mapped out the area between and around the conductors.7. With another color pen orpencil draw a point any place on your map torepresent a moveable positive charge. Predict the path it wouldtake bydrawing a line with your colored pen or pencil.8.
Now place a large metal ringin the tank between the electrodes. You willnotice that the potential will change as you move the probe inthe regionoutside of the ring but that it will remain the same when theprobe ismoved around in the region inside of the metal ring. Hence aconductor placed inan electric field will have the same potential everywhere.Summing Up:1. What generalizations ca nyou make from this exploration?2. Where would a positive testcharge have the least potential energy?3. How much energy must be addedto the system to move 1 electron, 1 meter in a direction alongone of the equal potential lines.4.
If lightning strikes a tree20 meters away, would it be better to stand facing thetree, your back to the tree, or your side to the tree? Assumeyour feet are acomfortable shoulder width apart. Explain your answer.Summing Up (Comments):1. Students' generalizationswill vary. They may recognize that equal potential linesare more concentrated near surfaces that have a smaller radiusof curvature. Somestudents may recognize that the force on a charge will be atright angles to theequal potential lines, but don't push this point until afterthe students have madetheir generalizations. This activity should be most appropriateto build the conceptof an electric field.2.
A positive test charge wouldhave the least potential energy next to the conductorestablished at 0 volts.3. If an electron is moved anydistance along the two volt equal potential line, nowork will be needed.4. If the person stood with theside facing the tree, the feet would be at differencepotentials and a dangerous shock could be the result.
If thefield around the tree isuniform, then facing or your back to the tree would place bothfeet on an equalpotential line and result in no shock. Facing away from the treewould probably bethe preferable option because the face would be protected fromflying debris andthe flash of light.Note: This experiment is notan electrostatic experiment. As electric current passesthrough the salt water, an IR drop is established and you areusing the voltmeter tomeasure this IR drop. As you move the probe, keeping the readingson thevoltmeter constant, you will be moving along regions of constantIR drop.Therefore, you will be identifying lines of constant potentialdifference. In thiselectric current situation, lines of constant potential differencecorrespond toequipotentials in the region surrounding the charges in an electrostaticsituation. Experiment # 2-Purpose:To show the electric field arounda simple charge configuration and to relate these electric linesto the equipotential lines in the previous experiment.Materials:This experiment usesthe same equipment as Experiment #1 on plotting equipotentialsexcept that the probe will now be in two contacts held a fixeddistance apart.
This can be accomplished with nails through asmall piece of wood or plastic (clear plastic will give a betterview). A separation between the nails of about 3 cm usually workswell.Procedure and Notes:Wire the apparatus exactly asin Experiment #1 except that this time the voltmeter will notbe attached to the '-' electrode, instead it will beacross the double probe. Place the double probe in the saltwater somewhere between the electrodes. Notice that, when yourotate the double probe, there will be a particular orientationwhich will give a maximum reading on the voltmeter. In this positionthe line between the two probes will indicate the direction ofthe electric field. If you start near the negative electrode,rotate the double probe until you find the maximum reading.
Thenmove it so that the first nail is now in the position of thesecond nail and again rotate to find the maximum reading. Thisprocess will map out the electric field between the two electrodes.Repeating this procedure with the map in another position willresult in plotting out another field line.
After you have plottedseveral field lines, compare these results with those of Experiment#1 and notice that theequipotentials are always perpendicular to the electric fieldlines.The apparatus is to be connectedto map out the electric field lines in the region between thetwo electrodes. Rotating the double probe to give a maximum voltagereading will indicate the direction of the electric field.1. Give specific values in volts of changes in electric potentialin the tray.2. Explain what is meant by theterm, 'Electric Field Lines'.When you move along anequipotential with a test charge, you will do no work. In anexternal electric field this would be possible only if the chargewere moved perpendicular to the field.
By the same token, ifyou move in a direction of maximum potential difference, youwill be moving along a electric field line and you do work. Rotatingthe double probe until the voltage across it is maximum resultsin finding the maximum potential change and, consequently, thedirection of the electric field.NOTE: This experiment is notan electrostatic experiment.
As electric currentpasses through the salt water, an IR drop is established andyou are using the volunteer to measure this IR drop.