#set document(title: "7.1 Electric Potential Energy", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 7.1#h(0.6em)Electric Potential Energy When a free positive charge #emph[q] is accelerated by an electric field, it is given kinetic energy . The process is analogous to an object being accelerated by a gravitational field, as if the charge were going down an electrical hill where its electric potential energy is converted into kinetic energy, although of course the sources of the forces are very different. Let us explore the work done on a charge #emph[q] by the electric field in this process, so that we may develop a definition of electric potential energy. #figure(figph[The first part of the figure shows two charged plates – one positive and one negative. A positive charge q is located between the plates and moves from point A to B. The second part of the figure shows a mass m rolling down a hill.], alt: "The first part of the figure shows two charged plates – one positive and one negative. A positive charge q is located between the plates and moves from point A to B. The second part of the figure shows a mass m rolling down a hill.", caption: [A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases, potential energy decreases as kinetic energy increases, #math.equation(block: false, alt: "minus Δ U equals Δ K")[$− Δ U = Δ K$]. Work is done by a force, but since this force is conservative, we can write #math.equation(block: false, alt: "W equals minus Δ U")[$W = − "Δ" U$].]) The electrostatic or Coulomb force is conservative, which means that the work done on #emph[q] is independent of the path taken, as we will demonstrate later. This is exactly analogous to the gravitational force. When a force is conservative, it is possible to define a potential energy associated with the force. It is usually easier to work with the potential energy (because it depends only on position) than to calculate the work directly. To show this explicitly, consider an electric charge #math.equation(block: false, alt: "plus q")[$+ q$] fixed at the origin and move another charge #math.equation(block: false, alt: "plus Q")[$+ Q$] toward #emph[q] in such a manner that, at each instant, the applied force #math.equation(block: false, alt: "F →")[$arrow("F")$] exactly balances the electric force #math.equation(block: false, alt: "F → sub e")[$arrow("F")_(e)$] on #emph[Q] . The work done by the applied force #math.equation(block: false, alt: "F →")[$arrow("F")$] on the charge #emph[Q] changes the potential energy of #emph[Q]. We call this potential energy the #strong[electrical potential energy] of #emph[Q]. #figure(figph[The figure shows two positive charges – fixed charge q and moving test charge Q and the forces on Q when is moved closer to q, from point P subscript 1 to point P subscript 2.], alt: "The figure shows two positive charges – fixed charge q and moving test charge Q and the forces on Q when is moved closer to q, from point P subscript 1 to point P subscript 2.", caption: [Displacement of “test” charge #emph[Q] in the presence of fixed “source” charge #emph[q].]) The work #math.equation(block: false, alt: "W sub 12")[$W_(12)$] done by the applied force #math.equation(block: false, alt: "F →")[$arrow("F")$] when the particle moves from #math.equation(block: false, alt: "P sub 1")[$P_(1)$] to #math.equation(block: false, alt: "P sub 2")[$P_(2)$] may be calculated by #math.equation(block: true, alt: "W sub 12 equals ∫ P sub 1 P sub 2 F → times d l → .")[$W_(12) = display(∫_(P_(1))^(P_(2)) arrow("F") · d arrow("l")) .$] Since the applied force #math.equation(block: false, alt: "F →")[$arrow("F")$] balances the electric force #math.equation(block: false, alt: "F → sub e")[$arrow("F")_(e)$] on #emph[Q], the two forces have equal magnitude and opposite directions. Therefore, the applied force is #math.equation(block: true, alt: "F → equals − F sub e → equals minus the fraction k sub e q Q over r squared r ^ ,")[$arrow("F") = "−" arrow(F_(e)) = − frac(k_(e) q Q, r^(2)) hat("r") ,$] where we have defined positive to be pointing away from the origin and #emph[r] is the distance from the origin. The directions of both the displacement and the applied force in the system in are parallel, and thus the work done on the system is positive. We use the letter #emph[U] to denote electric potential energy, which has units of joules (J). When a conservative force does negative work, the system gains potential energy. When a conservative force does positive work, the system loses potential energy, #math.equation(block: false, alt: "Δ U equals − W .")[$"Δ" U = "−" W .$] In the system in , the Coulomb force acts in the opposite direction to the displacement; therefore, the work is negative. However, we have increased the potential energy in the two-charge system. #examplebox("Example 1")[Kinetic Energy of a Charged Particle][ A #math.equation(block: false, alt: "plus 3.0 -nC")[$+ 3.0 "-nC"$] charge #emph[Q] is initially at rest a distance of 10 cm (#math.equation(block: false, alt: "r sub 1")[$r_(1)$]) from a #math.equation(block: false, alt: "plus 5.0 -nC")[$+ 5.0 "-nC"$] charge #emph[q] fixed at the origin . Naturally, the Coulomb force accelerates #emph[Q] away from #emph[q], eventually reaching 15 cm (#math.equation(block: false, alt: "r sub 2")[$r_(2)$]). #figure(figph[The figure shows two positive charges, q (+5.0nC) and Q (+3.0nC) and the repelling force on Q, marked as F subscript e. Q is located at r subscript 1 = 10cm and F subscript e vector is towards r subscript 2 = 15cm.], alt: "The figure shows two positive charges, q (+5.0nC) and Q (+3.0nC) and the repelling force on Q, marked as F subscript e. Q is located at r subscript 1 = 10cm and F subscript e vector is towards r subscript 2 = 15cm.", caption: [The charge #emph[Q] is repelled by #emph[q], thus having work done on it and gaining kinetic energy.]) + What is the work done by the electric field between #math.equation(block: false, alt: "r sub 1")[$r_(1)$] and #math.equation(block: false, alt: "r sub 2")[$r_(2)$]? + How much kinetic energy does #emph[Q] have at #math.equation(block: false, alt: "r sub 2")[$r_(2)$]? Strategy Calculate the work with the usual definition. Since #emph[Q] started from rest, this is the same as the kinetic energy. Solution Integrating force over distance, we obtain #math.equation(block: true, alt: "W sub 12, equals ∫ r sub 1 r sub 2 F → times d r → equals ∫ r sub 1 r sub 2 the fraction k sub e q Q over r squared d r equals [ minus the fraction k sub e q Q over r ] r sub 1 r sub 2 equals k sub e q Q [ the fraction −1 over r sub 2 plus the fraction 1 over r sub 1 ]; equals open parenthesis 8.99 times 10 to the power 9 Nm squared /C squared close parenthesis open parenthesis 5.0 times 10 to the power −9 C close parenthesis open parenthesis 3.0 times 10 to the power −9 C close parenthesis [ the fraction −1 over 0.15 m plus the fraction 1 over 0.10 m ]; equals 4.5 times 10 to the power −7 J .")[$W_(12) & = display(∫_(r_(1))^(r_(2)) arrow("F") · d arrow("r")) = display(∫_(r_(1))^(r_(2)) frac(k_(e) q Q, r^(2)) #h(0.2em) d r = attach([ − frac(k_(e) q Q, r) ], t: r_(2), b: r_(1))) = k_(e) q Q [ frac(−1, r_(2)) + frac(1, r_(1)) ] \ & = ( 8.99 #h(0.2em) × #h(0.2em) 10^(9) #h(0.2em) "Nm"^(2) "/C"^(2) ) ( 5.0 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "C" ) ( 3.0 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "C" ) [ frac(−1, 0.15 #h(0.2em) "m") + frac(1, 0.10 #h(0.2em) "m") ] \ & = 4.5 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "J" "."$] This is also the value of the kinetic energy at #math.equation(block: false, alt: "r sub 2 .")[$r_(2) .$] Significance Charge #emph[Q] was initially at rest; the electric field of #emph[q] did work on #emph[Q], so now #emph[Q] has kinetic energy equal to the work done by the electric field. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If #emph[Q] has a mass of #math.equation(block: false, alt: "4.00 μ g ,")[$4.00 #h(0.2em) μ "g" ,$] what is the speed of #emph[Q] at #math.equation(block: false, alt: "r sub 2 ?")[$r_(2) ?$] #solutionbox[ #math.equation(block: true, alt: "K equals the fraction 1 over 2 m v squared , v equals the square root of 2 the fraction K over m equals the square root of 2 the fraction 4.5 times 10 to the power −7 J over 4.00 times 10 to the power −9 kg equals 15 m/s")[$K = frac(1, 2) #h(0.2em) m v^(2) , #h(0.2em) v = sqrt(2 frac(K, m)) = sqrt(2 frac(4.5 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "J", 4.00 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "kg")) = 15 #h(0.2em) "m/s"$] ] ] In this example, the work #emph[W] done to accelerate a positive charge from rest is positive and results from a loss in #emph[U], or a negative #math.equation(block: false, alt: "Δ U")[$"Δ" U$]. A value for #emph[U] 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. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Electric Potential Energy] Work #emph[W] done to accelerate a positive charge from rest is positive and results from a loss in #emph[U], or a negative #math.equation(block: false, alt: "Δ U")[$"Δ" U$]. Mathematically, #math.equation(block: true, alt: "W equals − Δ U .")[$W = "−" "Δ" U .$] ] Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, for example, to use the concept of electric potential energy than to deal with the Coulomb force directly in real-world applications. In polar coordinates with #emph[q] at the origin and #emph[Q] located at #emph[r], the displacement element vector is #math.equation(block: false, alt: "d l → equals r ^ d r")[$d arrow("l") = hat("r") #h(0.2em) d r$] and thus the work becomes #math.equation(block: true, alt: "W sub 12 equals k sub e q Q ∫ r sub 1 r sub 2 the fraction 1 over r squared r ^ times r ^ d r = − k sub e q Q the fraction 1 over r sub 2 plus k sub e q Q the fraction 1 over r sub 1 .")[$W_(12) = k_(e) q Q display(∫_(r_(1))^(r_(2)) frac(1, r^(2)) hat("r") · hat("r") d r) = − k_(e) q Q frac(1, r_(2)) + k_(e) q Q frac(1, r_(1)) .$] Notice that this result only depends on the endpoints and is otherwise independent of the path taken. To explore this further, compare path #math.equation(block: false, alt: "P sub 1")[$P_(1)$] to #math.equation(block: false, alt: "P sub 2")[$P_(2)$] with path #math.equation(block: false, alt: "P sub 1 P sub 3 P sub 4 P sub 2")[$P_(1) P_(3) P_(4) P_(2)$] in . #figure(figph[The figure shows two positive charges, q and Q and the repelling force on Q. There are four points P subscript 1, P subscript 2, P subscript 3 and P subscript 4 where P subscript 1 P subscript 3 and P subscript 2 P subscript 4 form two concentric segments centered at q. The force on Q is perpendicular to direction of displacement when Q moves from P subscript 1 to P subscript 3 or P subscript 3 to P subscript 2.], alt: "The figure shows two positive charges, q and Q and the repelling force on Q. There are four points P subscript 1, P subscript 2, P subscript 3 and P subscript 4 where P subscript 1 P subscript 3 and P subscript 2 P subscript 4 form two concentric segments centered at q. The force on Q is perpendicular to direction of displacement when Q moves from P subscript 1 to P subscript 3 or P subscript 3 to P subscript 2.", caption: [Two paths for displacement #math.equation(block: false, alt: "P sub 1")[$P_(1)$] to #math.equation(block: false, alt: "P sub 2 .")[$P_(2) .$] The work on segments #math.equation(block: false, alt: "P sub 1 P sub 3")[$P_(1) P_(3)$] and #math.equation(block: false, alt: "P sub 4 P sub 2")[$P_(4) P_(2)$] are zero due to the electrical force being perpendicular to the displacement along these paths. Therefore, work on paths #math.equation(block: false, alt: "P sub 1 P sub 2")[$P_(1) P_(2)$] and #math.equation(block: false, alt: "P sub 1 P sub 3 P sub 4 P sub 2")[$P_(1) P_(3) P_(4) P_(2)$] are equal.]) The segments #math.equation(block: false, alt: "P sub 1 P sub 3")[$P_(1) P_(3)$] and #math.equation(block: false, alt: "P sub 4 P sub 2")[$P_(4) P_(2)$] are arcs of circles centered at #emph[q]. Since the force on #emph[Q] points either toward or away from #emph[q], no work is done by a force balancing the electric force, because it is perpendicular to the displacement along these arcs. Therefore, the only work done is along segment #math.equation(block: false, alt: "P sub 3 P sub 4 ,")[$P_(3) P_(4) ,$] which is identical to #math.equation(block: false, alt: "P sub 1 P sub 2 .")[$P_(1) P_(2) .$] One implication of this work calculation is that if we were to go around the path #math.equation(block: false, alt: "P sub 1 P sub 3 P sub 4 P sub 2 P sub 1 ,")[$P_(1) P_(3) P_(4) P_(2) P_(1) ,$] the net work would be zero . Recall that this is how we determine whether a force is conservative or not. Hence, because the electric force is related to the electric field by #math.equation(block: false, alt: "F → equals q E →")[$arrow("F") = q arrow("E")$], the electric field is itself conservative. That is, #math.equation(block: true, alt: "∮ E → times d l → equals 0 .")[$display(∮ arrow("E")) · d arrow("l") = 0 .$] Note that #emph[Q] is a constant. #figure(figph[The figure shows two positive charges, q and Q and the repelling force on Q. There are four points P subscript 1, P subscript 2, P subscript 3 and P subscript 4 where P subscript 1 P subscript 3 and P subscript 2 P subscript 4 form two concentric segments centered at q. Q’s direction of displacement is P subscript 1 P subscript 3 P subscript 4 P subscript 2 P subscript 1.], alt: "The figure shows two positive charges, q and Q and the repelling force on Q. There are four points P subscript 1, P subscript 2, P subscript 3 and P subscript 4 where P subscript 1 P subscript 3 and P subscript 2 P subscript 4 form two concentric segments centered at q. Q’s direction of displacement is P subscript 1 P subscript 3 P subscript 4 P subscript 2 P subscript 1.", caption: [A closed path in an electric field. The net work around this path is zero.]) Another implication is that we may define an electric potential energy. Recall that the work done by a conservative force is also expressed as the difference in the potential energy corresponding to that force. Therefore, the work #math.equation(block: false, alt: "W sub ref")[$W_("ref")$] to bring a charge from a reference point to a point of interest may be written as #math.equation(block: true, alt: "W sub ref equals ∫ r sub ref r F → times d l →")[$W_("ref") = display(∫_(r_("ref"))^(r) arrow("F") · d arrow("l"))$] and, by , the difference in potential energy #math.equation(block: false, alt: "open parenthesis U sub 2 minus U sub 1 close parenthesis")[$( U_(2) − U_(1) )$] of the test charge #emph[Q] between the two points is #math.equation(block: true, alt: "Δ U equals − ∫ r sub ref r F → times d l → .")[$"Δ" U = "−" display(∫_(r_("ref"))^(r) arrow("F") · d arrow("l")) .$] Therefore, we can write a general expression for the potential energy of two point charges (in spherical coordinates): #math.equation(block: true, alt: "Δ U equals − ∫ r sub ref r the fraction k sub e q Q over r squared d r equals − [ minus the fraction k sub e q Q over r ] r sub ref r equals k sub e q Q [ the fraction 1 over r minus the fraction 1 over r sub ref ] .")[$"Δ" U = "−" display(∫_(r_("ref"))^(r) frac(k_(e) q Q, r^(2)) d r = "−" attach([ − frac(k_(e) q Q, r) ], t: r, b: r_("ref"))) = k_(e) q Q [ frac(1, r) − frac(1, r_("ref")) ] .$] We may take the second term to be an arbitrary constant reference level, which serves as the zero reference: #math.equation(block: true, alt: "U open parenthesis r close parenthesis equals k sub e the fraction q Q over r minus U sub ref .")[$U ( r ) = k_(e) frac(q Q, r) − U_("ref") .$] A convenient choice of reference that relies on our common sense is that when the two charges are infinitely far apart, there is no interaction between them. (Recall the discussion of reference potential energy in Potential Energy and Conservation of Energy.) Taking the potential energy of this state to be zero removes the term #math.equation(block: false, alt: "U sub ref")[$U_("ref")$] from the equation (just like when we say the ground is zero potential energy in a gravitational potential energy problem), and the potential energy of #emph[Q] when it is separated from #emph[q] by a distance #emph[r] assumes the form #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "U open parenthesis r close parenthesis equals k sub e the fraction q Q over r open parenthesis z ero reference at r equals ∞ close parenthesis .")[$U ( r ) = k_(e) frac(q Q, r) #h(0.2em) ( z "ero reference at" #h(0.2em) r = ∞ ) .$] ] This formula is symmetrical with respect to #emph[q] and #emph[Q], so it is best described as the potential energy of the two-charge system. #examplebox("Example 2")[Potential Energy of a Charged Particle][ A #math.equation(block: false, alt: "plus 3.0 -nC")[$+ 3.0 "-nC"$] charge #emph[Q] is initially at rest a distance of 10 cm (#math.equation(block: false, alt: "r sub 1")[$r_(1)$]) from a #math.equation(block: false, alt: "+5 .0-nC")[$"+5" ".0-nC"$] charge #emph[q] fixed at the origin . Naturally, the Coulomb force accelerates #emph[Q] away from #emph[q], eventually reaching 15 cm (#math.equation(block: false, alt: "r sub 2")[$r_(2)$]). #figure(figph[The figure shows two positive charges, q (+5.0nC) and Q (+3.0nC) and the repelling force on Q, marked as F subscript e. Q is located at r subscript 1 = 10cm and F subscript e vector is towards r subscript 2 = 15cm.], alt: "The figure shows two positive charges, q (+5.0nC) and Q (+3.0nC) and the repelling force on Q, marked as F subscript e. Q is located at r subscript 1 = 10cm and F subscript e vector is towards r subscript 2 = 15cm.", caption: [The charge #emph[Q] is repelled by #emph[q], thus having work done on it and losing potential energy.]) What is the change in the potential energy of the two-charge system from #math.equation(block: false, alt: "r sub 1")[$r_(1)$] to #math.equation(block: false, alt: "r sub 2 ?")[$r_(2) ?$] Strategy Calculate the potential energy with the definition given above: #math.equation(block: false, alt: "Δ U sub 12 equals − ∫ r sub 1 r sub 2 F → times d r → .")[$"Δ" U_(12) = "−" display(∫_(r_(1))^(r_(2)) arrow("F") · d arrow("r")) .$] Since #emph[Q] started from rest, this is the same as the kinetic energy. Solution We have #math.equation(block: true, alt: "Δ U sub 12, equals − ∫ r sub 1 r sub 2 F → times d r → equals − ∫ r sub 1 r sub 2 the fraction k sub e q Q over r squared d r equals − [ minus the fraction k sub e q Q over r ] r sub 1 r sub 2 equals k sub e q Q [ the fraction 1 over r sub 2 minus the fraction 1 over r sub 1 ]; equals open parenthesis 8.99 times 10 to the power 9 Nm squared /C squared close parenthesis open parenthesis 5.0 times 10 to the power −9 C close parenthesis open parenthesis 3.0 times 10 to the power −9 C close parenthesis [ the fraction 1 over 0.15 m minus the fraction 1 over 0.10 m ]; equals −4.5 times 10 to the power −7 J .")[$"Δ" U_(12) & = "−" display(∫_(r_(1))^(r_(2)) arrow("F") · d arrow("r")) = "−" display(∫_(r_(1))^(r_(2)) frac(k_(e) q Q, r^(2)) d r = "−" attach([ − frac(k_(e) q Q, r) ], t: r_(2), b: r_(1))) = k_(e) q Q [ frac(1, r_(2)) − frac(1, r_(1)) ] \ & = ( 8.99 #h(0.2em) × #h(0.2em) 10^(9) #h(0.2em) "Nm"^(2) "/C"^(2) ) ( 5.0 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "C" ) ( 3.0 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "C" ) [ frac(1, 0.15 #h(0.2em) "m") − frac(1, 0.10 #h(0.2em) "m") ] \ & = −4.5 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "J" "."$] Significance The change in the potential energy is negative, as expected, and equal in magnitude to the change in kinetic energy in this system. Recall from that the change in kinetic energy was positive. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ What is the potential energy of #emph[Q] relative to the zero reference at infinity at #math.equation(block: false, alt: "r sub 2")[$r_(2)$] in the above example? #solutionbox[ It has kinetic energy of #math.equation(block: false, alt: "4.5 times 10 to the power −7 J")[$4.5 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "J"$] at point #math.equation(block: false, alt: "r sub 2")[$r_(2)$] and potential energy of #math.equation(block: false, alt: "9.0 times 10 to the power −7 J ,")[$9.0 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "J" ,$] which means that as #emph[Q] approaches infinity, its kinetic energy totals three times the kinetic energy at #math.equation(block: false, alt: "r sub 2 ,")[$r_(2) ,$] since all of the potential energy gets converted to kinetic. ] ] Due to Coulomb’s law, the forces due to multiple charges on a test charge #emph[Q] superimpose; they may be calculated individually and then added. This implies that the work integrals and hence the resulting potential energies exhibit the same behavior. To demonstrate this, we consider an example of assembling a system of four charges. #examplebox("Example 3")[Assembling Four Positive Charges][ Find the amount of work an external agent must do in assembling four charges #math.equation(block: false, alt: "plus 2.0 μ C ,")[$+ 2.0 #h(0.2em) μ "C" ,$]#math.equation(block: false, alt: "plus 3.0 μ C , plus 4.0 μ C ,")[$+ 3.0 #h(0.2em) μ "C" , #h(0.2em) + 4.0 #h(0.2em) μ "C" ,$] and #math.equation(block: false, alt: "plus 5.0 μ C")[$+ 5.0 #h(0.2em) μ "C"$] at the vertices of a square of side 1.0 cm, starting each charge from infinity . #figure(figph[The figure shows a square with side length 1.0cm and four charges (2.0µC, 3.0µC, 4.0µC and 5.0µC) located at four corners.], alt: "The figure shows a square with side length 1.0cm and four charges (2.0µC, 3.0µC, 4.0µC and 5.0µC) located at four corners.", caption: [How much work is needed to assemble this charge configuration?]) Strategy We bring in the charges one at a time, giving them starting locations at infinity and calculating the work to bring them in from infinity to their final location. We do this in order of increasing charge. Solution Step 1. First bring the #math.equation(block: false, alt: "plus 2.0 - μ C")[$+ 2.0 "-" μ "C"$] charge to the origin. Since there are no other charges at a finite distance from this charge yet, no work is done in bringing it from infinity, #math.equation(block: true, alt: "W sub 1 equals 0 .")[$W_(1) = 0 .$] Step 2. While keeping the #math.equation(block: false, alt: "plus 2.0 - μ C")[$+ 2.0 "-" μ "C"$] charge fixed at the origin, bring the #math.equation(block: false, alt: "plus 3.0 - μ C")[$+ 3.0 "-" μ "C"$] charge to #math.equation(block: false, alt: "open parenthesis x , y , z close parenthesis equals open parenthesis 1.0 cm , 0 , 0 close parenthesis")[$( x , y , z ) = ( 1.0 #h(0.2em) "cm" , 0 , 0 )$] . Now, the applied force must do work against the force exerted by the #math.equation(block: false, alt: "plus 2.0 - μ C")[$+ 2.0 "-" μ "C"$] charge fixed at the origin. The work done equals the change in the potential energy of the #math.equation(block: false, alt: "plus 3.0 - μ C")[$+ 3.0 "-" μ "C"$] charge: #math.equation(block: true, alt: "W sub 2 equals k sub e the fraction q sub 1 q sub 2 over r sub 12 equals open parenthesis 9.0 times 10 to the power 9 the fraction N times m squared over C squared close parenthesis the fraction open parenthesis 2.0 times 10 to the power −6 C close parenthesis open parenthesis 3.0 times 10 to the power −6 C close parenthesis over 1.0 times 10 to the power −2 m equals 5.4 J .")[$W_(2) = k_(e) frac(q_(1) q_(2), r_(12)) = ( 9.0 #h(0.2em) × #h(0.2em) 10^(9) frac("N" · "m"^(2), "C"^(2)) ) frac(( 2.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ) ( 3.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ), 1.0 #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "m") = 5.4 #h(0.2em) "J" "."$] #figure(figph[The figure shows a square with side length 1.0cm and two charges (2.0µC and 3.0µC) on adjacent corners.], alt: "The figure shows a square with side length 1.0cm and two charges (2.0µC and 3.0µC) on adjacent corners.", caption: [Step 2. Work #math.equation(block: false, alt: "W sub 2")[$W_(2)$] to bring the #math.equation(block: false, alt: "plus 3.0 - μ C")[$+ 3.0 "-" μ "C"$] charge from infinity.]) Step 3. While keeping the charges of #math.equation(block: false, alt: "plus 2.0 μ C")[$+ 2.0 #h(0.2em) μ "C"$] and #math.equation(block: false, alt: "plus 3.0 μ C")[$+ 3.0 #h(0.2em) μ "C"$] fixed in their places, bring in the #math.equation(block: false, alt: "plus 4.0 - μ C")[$+ 4.0 "-" μ "C"$] charge to #math.equation(block: false, alt: "open parenthesis x , y , z close parenthesis equals open parenthesis 1.0 cm , 1.0 cm , 0 close parenthesis")[$( x , y , z ) = ( 1.0 #h(0.2em) "cm" , 1.0 #h(0.2em) "cm" , 0 )$] . The work done in this step is #math.equation(block: true, alt: "W sub 3, equals k sub e the fraction q sub 1 q sub 3 over r sub 13 plus k sub e the fraction q sub 2 q sub 3 over r sub 23; equals open parenthesis 9.0 times 10 to the power 9 the fraction N times m squared over C squared close parenthesis [ the fraction open parenthesis 2.0 times 10 to the power −6 C close parenthesis open parenthesis 4.0 times 10 to the power −6 C close parenthesis over the square root of 2 times 10 to the power −2 m plus the fraction open parenthesis 3.0 times 10 to the power −6 C close parenthesis open parenthesis 4.0 times 10 to the power −6 C close parenthesis over 1.0 times 10 to the power −2 m ] equals 15.9 J .")[$W_(3) & = k_(e) frac(q_(1) q_(3), r_(13)) + k_(e) frac(q_(2) q_(3), r_(23)) \ & = ( 9.0 #h(0.2em) × #h(0.2em) 10^(9) frac("N" · "m"^(2), "C"^(2)) ) [ frac(( 2.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ) ( 4.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ), sqrt(2) #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "m") + frac(( 3.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ) ( 4.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ), 1.0 #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "m") ] = 15.9 #h(0.2em) "J" "."$] #figure(figph[The figure shows a square with side length 1.0cm and three charges (2.0µC, 3.0µC and 4.0µC) on three corners.], alt: "The figure shows a square with side length 1.0cm and three charges (2.0µC, 3.0µC and 4.0µC) on three corners.", caption: [Step 3. The work #math.equation(block: false, alt: "W sub 3")[$W_(3)$] to bring the #math.equation(block: false, alt: "plus 4.0 - μ C")[$+ 4.0 "-" μ "C"$] charge from infinity.]) Step 4. Finally, while keeping the first three charges in their places, bring the #math.equation(block: false, alt: "plus 5.0 - μ C")[$+ 5.0 "-" μ "C"$] charge to #math.equation(block: false, alt: "open parenthesis x , y , z close parenthesis equals open parenthesis 0 , 1.0 cm , 0 close parenthesis")[$( x , y , z ) = ( 0 , #h(0.2em) 1.0 #h(0.2em) "cm" , #h(0.2em) 0 )$] . The work done here is #math.equation(block: true, alt: "W sub 4, equals k sub e q sub 4 [ the fraction q sub 1 over r sub 14 plus the fraction q sub 2 over r sub 24 plus the fraction q sub 3 over r sub 34 ] ,; equals open parenthesis 9.0 times 10 to the power 9 the fraction N times m squared over C squared close parenthesis open parenthesis 5.0 times 10 to the power −6 C close parenthesis [ the fraction open parenthesis 2.0 times 10 to the power −6 C close parenthesis over 1.0 times 10 to the power −2 m plus the fraction open parenthesis 3.0 times 10 to the power −6 C close parenthesis over the square root of 2 times 10 to the power −2 m plus the fraction open parenthesis 4.0 times 10 to the power −6 C close parenthesis over 1.0 times 10 to the power −2 m ] equals 36.5 J .")[$W_(4) & = k_(e) q_(4) [ frac(q_(1), r_(14)) + frac(q_(2), r_(24)) + frac(q_(3), r_(34)) ] , \ & = ( 9.0 #h(0.2em) × #h(0.2em) 10^(9) frac("N" · "m"^(2), "C"^(2)) ) ( 5.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ) [ frac(( 2.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ), 1.0 #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "m") + frac(( 3.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ), sqrt(2) #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "m") + frac(( 4.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C" ), 1.0 #h(0.2em) × #h(0.2em) 10^(−2) #h(0.2em) "m") ] = 36.5 #h(0.2em) "J" "."$] #figure(figph[The figure shows a square with side length 1.0cm and four charges (2.0µC, 3.0µC, 4.0µC and 5.0µC) located at four corners.], alt: "The figure shows a square with side length 1.0cm and four charges (2.0µC, 3.0µC, 4.0µC and 5.0µC) located at four corners.", caption: [Step 4. The work #math.equation(block: false, alt: "W sub 4")[$W_(4)$] to bring the #math.equation(block: false, alt: "plus 5.0 - μ C")[$+ 5.0 "-" μ "C"$] charge from infinity.]) Hence, the total work done by the applied force in assembling the four charges is equal to the sum of the work in bringing each charge from infinity to its final position: #math.equation(block: true, alt: "W sub T equals W sub 1 plus W sub 2 plus W sub 3 plus W sub 4 equals 0 plus 5.4 J plus 15.9 J plus 36.5 J equals 57.8 J .")[$W_("T") = W_(1) + W_(2) + W_(3) + W_(4) = 0 + 5.4 #h(0.2em) "J" + 15.9 #h(0.2em) "J" + 36.5 #h(0.2em) "J" = 57.8 #h(0.2em) "J" "."$] Significance The work on each charge depends only on its pairwise interactions with the other charges. No more complicated interactions need to be considered; the work on the third charge only depends on its interaction with the first and second charges, the interaction between the first and second charge does not affect the third. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is the electrical potential energy of two point charges positive or negative if the charges are of the same sign? Opposite signs? How does this relate to the work necessary to bring the charges into proximity from infinity? #solutionbox[ positive, negative, and these quantities are the same as the work you would need to do to bring the charges in from infinity ] ] Note that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, and negative if the two charges are of opposite types. This makes sense if you think of the change in the potential energy #math.equation(block: false, alt: "Δ U")[$"Δ" U$] as you bring the two charges closer or move them farther apart. Depending on the relative types of charges, you may have to work on the system or the system would do work on you, that is, your work is either positive or negative. If you have to do positive work on the system (actually push the charges closer), then the energy of the system should increase. If you bring two positive charges or two negative charges closer, you have to do positive work on the system, which raises their potential energy. Since potential energy is proportional to 1/#emph[r], the potential energy goes up when #emph[r] goes down between two positive or two negative charges. On the other hand, if you bring a positive and a negative charge nearer, you have to do negative work on the system (the charges are pulling you), which means that you take energy away from the system. This reduces the potential energy. Since potential energy is negative in the case of a positive and a negative charge pair, the increase in 1/#emph[r] makes the potential energy more negative, which is the same as a reduction in potential energy. The result from may be extended to systems with any arbitrary number of charges. In this case, it is most convenient to write the formula as #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "W sub 12 ⋯ N equals the fraction k sub e over 2 ∑ i N ∑ j N the fraction q sub i q sub j over r sub i j for i not equal to j .")[$W_(12 ⋯ N) = frac(k_(e), 2) display(limits(∑)^(N)_(i) display(limits(∑)^(N)_(j) frac(q_(i) q_(j), r_(i j)))) #h(0.2em) "for" #h(0.2em) i ≠ j .$] ] The factor of 1/2 accounts for adding each pair of charges twice. === Summary - The work done to move a charge from point #emph[A] to #emph[B] in an electric field is path independent, and the work around a closed path is zero. Therefore, the electric field and electric force are conservative. - We can define an electric potential energy, which between point charges is #math.equation(block: false, alt: "U open parenthesis r close parenthesis equals k sub e the fraction q Q over r")[$U ( r ) = k_(e) frac(q Q, r)$], with the zero reference taken to be at infinity. - The superposition principle holds for electric potential energy; the potential energy of a system of multiple charges is the sum of the potential energies of the individual pairs. === Conceptual Questions Would electric potential energy be meaningful if the electric field were not conservative? #solutionbox[ No. We can only define potential energies for conservative fields. ] Why do we need to be careful about work done #emph[on] the system versus work done #emph[by] the system in calculations? Does the order in which we assemble a system of point charges affect the total work done? #solutionbox[ No, though certain orderings may be simpler to compute. ] === Problems Consider a charge #math.equation(block: false, alt: "Q sub 1 open parenthesis plus 5.0 μ C close parenthesis")[$Q_(1) ( + 5.0 #h(0.2em) μ "C" )$] fixed at a site with another charge #math.equation(block: false, alt: "Q sub 2")[$Q_(2)$] (charge #math.equation(block: false, alt: "plus 3.0 μ C")[$+ 3.0 #h(0.2em) μ "C"$], mass #math.equation(block: false, alt: "6.0 μ g close parenthesis")[$6.0 #h(0.2em) μ "g" \)$] moving in the neighboring space. (a) Evaluate the potential energy of #math.equation(block: false, alt: "Q sub 2")[$Q_(2)$] when it is 4.0 cm from #math.equation(block: false, alt: "Q sub 1 .")[$Q_(1) .$] (b) If #math.equation(block: false, alt: "Q sub 2")[$Q_(2)$] starts from rest from a point 4.0 cm from #math.equation(block: false, alt: "Q sub 1 ,")[$Q_(1) ,$] what will be its speed when it is 8.0 cm from #math.equation(block: false, alt: "Q sub 1")[$Q_(1)$]? (#emph[Note:] #math.equation(block: false, alt: "Q sub 1")[$Q_(1)$] is held fixed in its place.) #solutionbox[ a. #math.equation(block: false, alt: "U equals 3.4 J;")[$U = 3.4 #h(0.2em) "J;"$] #linebreak() b. #math.equation(block: false, alt: "the fraction 1 over 2 m v squared equals k Q sub 1 Q sub 2 open parenthesis the fraction 1 over r sub i minus the fraction 1 over r sub f close parenthesis → v equals 2.4 x 10 to the power 4 m/s")[$frac(1, 2) m v^(2) = k Q_(1) Q_(2) ( frac(1, r_(i)) − frac(1, r_(f)) ) → v = 2.4 x 10^(4) #h(0.2em) "m/s"$] ] Two charges #math.equation(block: false, alt: "Q sub 1 open parenthesis plus 2.00 μ C close parenthesis")[$Q_(1) ( + 2.00 #h(0.2em) μ "C" )$] and #math.equation(block: false, alt: "Q sub 2 open parenthesis plus 2.00 μ C close parenthesis")[$Q_(2) ( + 2.00 #h(0.2em) μ "C" )$] are placed symmetrically along the #emph[x]-axis at #math.equation(block: false, alt: "x equals plus or minus 3.00 cm")[$x = ± 3.00 #h(0.2em) "cm"$]. Consider a charge #math.equation(block: false, alt: "Q sub 3")[$Q_(3)$] of charge #math.equation(block: false, alt: "plus 4.00 μ C")[$+ 4.00 #h(0.2em) μ "C"$] and mass 10.0 mg moving along the #emph[y]-axis. If #math.equation(block: false, alt: "Q sub 3")[$Q_(3)$] starts from rest at #math.equation(block: false, alt: "y equals 2.00 cm,")[$y = 2.00 #h(0.2em) "cm,"$] what is its speed when it reaches #math.equation(block: false, alt: "y equals 4.00 cm?")[$y = 4.00 #h(0.2em) "cm?"$] To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of #math.equation(block: false, alt: "0.529 times 10 to the power −10 m,")[$0.529 #h(0.2em) × #h(0.2em) 10^(−10) #h(0.2em) "m,"$] the average distance between proton and electron in a hydrogen atom. How much work is done? #solutionbox[ #math.equation(block: true, alt: "U equals 4.36 times 10 to the power −18 J")[$U = 4.36 #h(0.2em) × #h(0.2em) 10^(−18) #h(0.2em) "J" #h(0.2em)$] ] (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?