#set document(title: "5.4 Electric Field", 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")) == 5.4#h(0.6em)Electric Field As we showed in the preceding section, the net electric force on a test charge is the vector sum of all the electric forces acting on it, from all of the various source charges, located at their various positions. But what if we use a different test charge, one with a different magnitude, or sign, or both? Or suppose we have a dozen different test charges we wish to try at the same location? We would have to calculate the sum of the forces from scratch. Fortunately, it is possible to define a quantity, called the #strong[electric field], which is independent of the test charge. It only depends on the configuration of the source charges, and once found, allows us to calculate the force on any test charge. === Defining a Field Suppose we have #emph[N] source charges #math.equation(block: false, alt: "q sub 1 , q sub 2 , q sub 3 ,… , q sub N")[$q_(1) , q_(2) , q_(3) ",…" , q_(N)$], applying #emph[N] electrostatic forces on a test charge #emph[Q], at displacements #math.equation(block: false, alt: "r → sub 1 , r → sub 2 , r → sub 3 ,… , r → sub N")[$arrow("r")_(1) , arrow("r")_(2) , arrow("r")_(3) ",…" , arrow("r")_(N)$] from the source charges. The net force on #emph[Q] is #math.equation(block: true, alt: "F →, equals F → sub 1 plus F → sub 2 plus F → sub 3 plus ⋯ plus F → sub N; equals the fraction 1 over 4 π ε sub 0 open parenthesis the fraction Q q sub 1 over r 1 2 r ^ sub 1 plus the fraction Q q sub 2 over r 2 2 r ^ sub 2 plus the fraction Q q sub 3 over r 3 2 r ^ sub 3 plus ⋯ plus the fraction Q q sub N over r N 2 r ^ sub N close parenthesis; equals Q [ the fraction 1 over 4 π ε sub 0 open parenthesis the fraction q sub 1 over r 1 2 r ^ sub 1 plus the fraction q sub 2 over r 2 2 r ^ sub 2 plus the fraction q sub 3 over r 3 2 r ^ sub 3 plus ⋯ plus the fraction q sub N over r N 2 r ^ sub N close parenthesis ] .")[$arrow("F") & = arrow("F")_(1) + arrow("F")_(2) + arrow("F")_(3) + ⋯ + arrow("F")_(N) \ & = frac(1, 4 π ε_(0)) ( frac(Q q_(1), r_(1)^(2)) hat("r")_(1) + frac(Q q_(2), r_(2)^(2)) hat("r")_(2) + frac(Q q_(3), r_(3)^(2)) hat("r")_(3) + ⋯ + frac(Q q_(N), r_(N)^(2)) hat("r")_(N) ) \ & = Q [ frac(1, 4 π ε_(0)) ( frac(q_(1), r_(1)^(2)) hat("r")_(1) + frac(q_(2), r_(2)^(2)) hat("r")_(2) + frac(q_(3), r_(3)^(2)) hat("r")_(3) + ⋯ + frac(q_(N), r_(N)^(2)) hat("r")_(N) ) ] .$] We can rewrite this as #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "F → equals Q E →")[$arrow("F") = Q arrow("E")$] ] where #math.equation(block: true, alt: "E → ≡ the fraction 1 over 4 π ε sub 0 open parenthesis the fraction q sub 1 over r 1 2 r ^ sub 1 plus the fraction q sub 2 over r 2 2 r ^ sub 2 plus the fraction q sub 3 over r 3 2 r ^ sub 3 plus ⋯ plus the fraction q sub N over r N 2 r ^ sub N close parenthesis")[$arrow("E") ≡ frac(1, 4 π ε_(0)) ( frac(q_(1), r_(1)^(2)) hat("r")_(1) + frac(q_(2), r_(2)^(2)) hat("r")_(2) + frac(q_(3), r_(3)^(2)) hat("r")_(3) + ⋯ + frac(q_(N), r_(N)^(2)) hat("r")_(N) )$] or, more compactly, #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "E → open parenthesis P close parenthesis ≡ the fraction 1 over 4 π ε sub 0 ∑ i equals 1 N the fraction q sub i over r i 2 r ^ sub i .")[$arrow("E") ( P ) ≡ frac(1, 4 π ε_(0)) display(limits(∑)^(N)_(i = 1) frac(q_(i), r_(i)^(2))) hat("r")_(i) .$] ] This expression is called the electric field at position #math.equation(block: false, alt: "P equals P open parenthesis x , y , z close parenthesis")[$P = P ( x , y , z )$] of the #emph[N] source charges. Here, #emph[P] is the location of the point in space where you are calculating the field and the vectors #math.equation(block: false, alt: "r → sub i")[$arrow("r")_(i)$] are the displacements from each source to the test point . Note that we have to impose a coordinate system to solve actual problems. #figure(figph[Five source charges are shown as small spheres distributed within an x y z coordinate system. The sources are labeled q sub 1, q sub 2, and so on. Sources 1, 2, and 4 are shaded red and sources 3 and 5 are shaded blue. A test point is also shown and labeled as point P. The electric field vectors due to each source are shown as an arrow at point P, pointing toward point P and labeled with the index of the associated source. Vector E 1 points away from q 1, E 2 away from q 2, and E 4 away from q 4. Vector E 3 points toward q 3 and vector E 5 toward q 5.], alt: "Five source charges are shown as small spheres distributed within an x y z coordinate system. The sources are labeled q sub 1, q sub 2, and so on. Sources 1, 2, and 4 are shaded red and sources 3 and 5 are shaded blue. A test point is also shown and labeled as point P. The electric field vectors due to each source are shown as an arrow at point P, pointing toward point P and labeled with the index of the associated source. Vector E 1 points away from q 1, E 2 away from q 2, and E 4 away from q 4. Vector E 3 points toward q 3 and vector E 5 toward q 5.", caption: [Each of these five source charges creates its own electric field at every point in space; shown here are the field vectors at an arbitrary point #emph[P]. Like the electric force, the net electric field obeys the superposition principle.]) Notice that the calculation of the electric field makes no reference to the test charge. Thus, the physically useful approach is to calculate the electric field and then use it to calculate the force on some test charge later, if needed. Different test charges experience different forces , but it is the same electric field . That being said, recall that there is no fundamental difference between a test charge and a source charge; these are merely convenient labels for the system of interest. Any charge produces an electric field; however, just as Earth’s orbit is not affected by Earth’s own gravity, a charge is not subject to a force due to the electric field it generates. Charges are only subject to forces from the electric fields of other charges. In this respect, the electric field #math.equation(block: false, alt: "E →")[$arrow("E")$] of a point charge is similar to the gravitational field #math.equation(block: false, alt: "g →")[$arrow(g)$] of Earth; once we have calculated the gravitational field at some point in space, we can use it any time we want to calculate the resulting force on any mass we choose to place at that point. In fact, this is exactly what we do when we say the gravitational field of Earth (near Earth’s surface) has a value of #math.equation(block: false, alt: "9.81 m/s squared ,")[$9.81 #h(0.2em) "m/s"^(2) ,$] and then we calculate the resulting force (i.e., weight) on different masses. Also, the general expression for calculating #math.equation(block: false, alt: "g →")[$arrow(g)$] at arbitrary distances from the center of Earth (i.e., not just near Earth’s surface) is very similar to the expression for #math.equation(block: false, alt: "E →")[$arrow("E")$]: #math.equation(block: false, alt: "g → equals G the fraction M over r squared r ^")[$arrow(g) = G frac(M, r^(2)) hat("r")$], where #emph[G] is a proportionality constant, playing the same role for #math.equation(block: false, alt: "g →")[$arrow(g)$] as #math.equation(block: false, alt: "the fraction 1 over 4 π ε sub 0")[$frac(1, 4 π ε_(0))$] does for #math.equation(block: false, alt: "E →")[$arrow("E")$]. The value of #math.equation(block: false, alt: "g →")[$arrow(g)$] is calculated once and is then used in an endless number of problems. To push the analogy further, notice the units of the electric field: From #math.equation(block: false, alt: "F equals Q E")[$F = Q E$], the units of #emph[E] are newtons per coulomb, N/C, that is, the electric field applies a force on each unit charge. Now notice the units of #emph[g]: From #math.equation(block: false, alt: "w equals m g")[$w = m g$], the units of #emph[g] are newtons per kilogram, N/kg, that is, the gravitational field applies a force on each unit mass. We could say that the gravitational field of Earth, near Earth’s surface, has a value of 9.81 N/kg. === The Meaning of “Field” Recall from your studies of gravity that the word “field” in this context has a precise meaning. A field, in physics, is a physical quantity whose value depends on (is a function of) position, relative to the source of the field. In the case of the electric field, shows that the value of #math.equation(block: false, alt: "E →")[$arrow("E")$] (both the magnitude and the direction) depends on where in space the point #emph[P] is located, with #math.equation(block: false, alt: "r → sub i")[$arrow("r")_(i)$] measured from the locations of the source charges #math.equation(block: false, alt: "q sub i")[$q_(i)$]. In addition, since the electric field is a vector quantity, the electric field is referred to as a #strong[#emph[vector field]]. (The gravitational field is also a vector field.) In contrast, a field that has only a magnitude at every point is a #strong[#emph[scalar field]]. The temperature in a room is an example of a scalar field. It is a field because the temperature, in general, is different at different locations in the room, and it is a scalar field because temperature is a scalar quantity. Also, as you did with the gravitational field of an object with mass, you should picture the electric field of a charge-bearing object (the source charge) as a continuous, immaterial substance that surrounds the source charge, filling all of space—in principle, to #math.equation(block: false, alt: "± ∞")[$"±" ∞$] in all directions. The field exists at every physical point in space. To put it another way, the electric charge on an object alters the space around the charged object in such a way that all other electrically charged objects in space experience an electric force as a result of being in that field. The electric field, then, is the mechanism by which the electric properties of the source charge are transmitted to and through the rest of the universe. (Again, the range of the electric force is infinite.) We will see in subsequent chapters that the speed at which electrical phenomena travel is the same as the speed of light. There is a deep connection between the electric field and light. === Superposition Yet another experimental fact about the field is that it obeys the superposition principle. In this context, that means that we can (in principle) calculate the total electric field of many source charges by calculating the electric field of only #math.equation(block: false, alt: "q sub 1")[$q_(1)$] at position #emph[P], then calculate the field of #math.equation(block: false, alt: "q sub 2")[$q_(2)$] at #emph[P], while—and this is the crucial idea—ignoring the field of, and indeed even the existence of, #math.equation(block: false, alt: "q sub 1 .")[$q_(1) .$] We can repeat this process, calculating the field of each individual source charge, independently of the existence of any of the other charges. The total electric field, then, is the vector sum of all these fields. That, in essence, is what says. In the next section, we describe how to determine the shape of an electric field of a source charge distribution and how to sketch it. === The Direction of the Field enables us to determine the magnitude of the electric field, but we need the direction also. We use the convention that the direction of any electric field vector is the same as the direction of the electric force vector that the field would apply to a positive test charge placed in that field. Such a charge would be repelled by positive source charges (the force on it would point away from the positive source charge) but attracted to negative charges (the force points toward the negative source). #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Direction of the Electric Field] By convention, all electric fields #math.equation(block: false, alt: "E →")[$arrow("E")$] point away from positive source charges and point toward negative source charges. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Add charges to the #link("https://openstax.org/l/21elefiedream")[Electric Field of Dreams] and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude. ] #examplebox("Example 1")[The #emph[E]-Field of an Atom][ In an ionized helium atom, the most probable distance between the nucleus and the electron is #math.equation(block: false, alt: "r equals 26.5 times 10 to the power −12 m")[$r = 26.5 #h(0.2em) × #h(0.2em) 10^(−12) #h(0.2em) "m"$]. What is the electric field due to the nucleus at the location of the electron? Strategy Note that although the electron is mentioned, it is not used in any calculation. The problem asks for an electric field, not a force; hence, there is only one charge involved, and the problem specifically asks for the field due to the nucleus. Thus, the electron is a red herring; only its distance matters. Also, since the distance between the two protons in the nucleus is much, much smaller than the distance of the electron from the nucleus, we can treat the two protons as a single charge +2#emph[e] . #figure(figph[A positive charge of plus 2 e is shown at the center of a sphere of radius r. An electron is depicted as a particle on the sphere. The vector r is shown as a vector with its tail at the center and its head at the location of the electron. The electric field at the location of the electron is shown as a vector E with its tail at the electron and pointing directly away from the center.], alt: "A positive charge of plus 2 e is shown at the center of a sphere of radius r. An electron is depicted as a particle on the sphere. The vector r is shown as a vector with its tail at the center and its head at the location of the electron. The electric field at the location of the electron is shown as a vector E with its tail at the electron and pointing directly away from the center.", caption: [A schematic representation of a helium atom. Again, helium physically looks nothing like this, but this sort of diagram is helpful for calculating the electric field of the nucleus.]) Solution The electric field is calculated by #math.equation(block: true, alt: "E → equals the fraction 1 over 4 π ε sub 0 ∑ i equals 1 N the fraction q sub i over r i 2 r ^ sub i .")[$arrow("E") = frac(1, 4 π ε_(0)) display(limits(∑)^(N)_(i = 1) frac(q_(i), r_(i)^(2))) hat("r")_(i) .$] Since there is only one source charge (the nucleus), this expression simplifies to #math.equation(block: true, alt: "E → equals the fraction 1 over 4 π ε sub 0 the fraction q over r squared r ^ .")[$arrow("E") = frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) hat(r) .$] Here #math.equation(block: false, alt: "q equals 2 e equals 2 open parenthesis 1.6 times 10 to the power −19 C close parenthesis")[$q = 2 e = 2 ( 1.6 #h(0.2em) × #h(0.2em) 10^(−19) #h(0.2em) "C" )$] (since there are two protons) and #emph[r] is given; substituting gives #math.equation(block: true, alt: "E → equals the fraction 1 over 4 π open parenthesis 8.85 times 10 to the power −12 the fraction C squared over N times m squared close parenthesis the fraction 2 open parenthesis 1.6 times 10 to the power −19 C close parenthesis over open parenthesis 26.5 times 10 to the power −12 m close parenthesis squared r ^ equals 4.1 times 10 to the power 12 the fraction N over C r ^ .")[$arrow("E") = frac(1, 4 π ( 8.85 #h(0.2em) × #h(0.2em) 10^(−12) frac("C"^(2), "N" · "m"^(2)) )) #h(0.2em) frac(2 ( 1.6 #h(0.2em) × #h(0.2em) 10^(−19) #h(0.2em) "C" ), attach(( 26.5 #h(0.2em) × #h(0.2em) 10^(−12) #h(0.2em) "m" ), t: 2)) hat("r") = 4.1 #h(0.2em) × #h(0.2em) 10^(12) frac("N", "C") hat("r") .$] The direction of #math.equation(block: false, alt: "E →")[$arrow("E")$] is radially away from the nucleus in all directions. Why? Because a positive test charge placed in this field would accelerate radially away from the nucleus (since it is also positively charged), and again, the convention is that the direction of the electric field vector is defined in terms of the direction of the force it would apply to positive test charges. ] #examplebox("Example 2")[The #emph[E]-Field above Two Equal Charges][ (a) Find the electric field (magnitude and direction) a distance #emph[z] above the midpoint between two equal charges #math.equation(block: false, alt: "+ q")[$"+" q$] that are a distance #emph[d] apart . Check that your result is consistent with what you’d expect when #math.equation(block: false, alt: "z ≫ d")[$z ≫ d$]. (b) The same as part (a), only this time make the right-hand charge #math.equation(block: false, alt: "− q")[$"−" q$] instead of #math.equation(block: false, alt: "+ q")[$"+" q$]. #figure(figph[Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta.], alt: "Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta.", caption: [Finding the field of two identical source charges at the point #emph[P]. Due to the symmetry, the net field at #emph[P] is entirely vertical. (Notice that this is #emph[not] true away from the midline between the charges.)]) Strategy We add the two fields as vectors, per . Notice that the system (and therefore the field) is symmetrical about the vertical axis; as a result, the horizontal components of the field vectors cancel. This simplifies the math. Also, we take care to express our final answer in terms of only quantities that are given in the original statement of the problem: #emph[q], #emph[z], #emph[d], and constants #math.equation(block: false, alt: "open parenthesis π , ε sub 0 close parenthesis .")[$( π , ε_(0) ) .$] Solution + By symmetry, the horizontal (#emph[x])-components of #math.equation(block: false, alt: "E →")[$arrow("E")$] cancel ; #math.equation(block: false, alt: "E sub x equals the fraction 1 over 4 π ε sub 0 the fraction q over r squared sin θ minus the fraction 1 over 4 π ε sub 0 the fraction q over r squared sin θ equals 0")[$E_(x) = frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) #h(0.2em) "sin" #h(0.2em) θ − frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) #h(0.2em) "sin" #h(0.2em) θ = 0$]. #linebreak() #figure(figph[Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta. The x and y components of the electric field are shown as arrows whose tails are at point P. Four arrows are shown, as follows: E sub x r points to the left, E sub x l points to the right, E sub z l points up, and E sub z r points up.], alt: "Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta. The x and y components of the electric field are shown as arrows whose tails are at point P. Four arrows are shown, as follows: E sub x r points to the left, E sub x l points to the right, E sub z l points up, and E sub z r points up.", caption: [Note that the horizontal components of the electric fields from the two charges cancel each other out, while the vertical components add together.]) #linebreak() The vertical (#emph[z])-component is given by #math.equation(block: true, alt: "E sub z equals the fraction 1 over 4 π ε sub 0 the fraction q over r squared cos θ plus the fraction 1 over 4 π ε sub 0 the fraction q over r squared cos θ equals the fraction 1 over 4 π ε sub 0 the fraction 2 q over r squared cos θ .")[$E_(z) = frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) #h(0.2em) "cos" #h(0.2em) θ + frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) #h(0.2em) "cos" #h(0.2em) θ = frac(1, 4 π ε_(0)) #h(0.2em) frac(2 q, r^(2)) #h(0.2em) "cos" #h(0.2em) θ .$] Since none of the other components survive, this is the entire electric field, and it points in the #math.equation(block: false, alt: "k ^")[$hat("k")$] direction. Notice that this calculation uses the principle of #strong[superposition]; we calculate the fields of the two charges independently and then add them together. #linebreak() What we want to do now is replace the quantities in this expression that we don’t know (such as #emph[r]), or can’t easily measure (such as #math.equation(block: false, alt: "cos θ close parenthesis")[$"cos" #h(0.2em) θ \)$] with quantities that we do know, or can measure. In this case, by geometry, #math.equation(block: true, alt: "r squared equals z squared plus open parenthesis the fraction d over 2 close parenthesis squared")[$r^(2) = z^(2) + attach(( frac(d, 2) ), t: 2)$] and #math.equation(block: true, alt: "cos θ equals the fraction z over r equals the fraction z over [ z squared plus open parenthesis the fraction d over 2 close parenthesis squared ] to the power 1 / 2 .")[$"cos" #h(0.2em) θ = frac(z, r) = frac(z, attach([ z^(2) + attach(( frac(d, 2) ), t: 2) ], t: 1 "/" 2)) .$] Thus, substituting, #math.equation(block: true, alt: "E → open parenthesis z close parenthesis equals the fraction 1 over 4 π ε sub 0 the fraction 2 q over [ z squared plus open parenthesis the fraction d over 2 close parenthesis squared ] the fraction z over [ z squared plus open parenthesis the fraction d over 2 close parenthesis squared ] to the power 1 / 2 k ^ .")[$arrow("E") ( z ) = frac(1, 4 π ε_(0)) #h(0.2em) frac(2 q, [ z^(2) + attach(( frac(d, 2) ), t: 2) ]) #h(0.2em) frac(z, attach([ z^(2) + attach(( frac(d, 2) ), t: 2) ], t: 1 "/" 2)) hat("k") .$] Simplifying, the desired answer is #math.equation(block: true, alt: "E → open parenthesis z close parenthesis equals the fraction 1 over 4 π ε sub 0 the fraction 2 q z over [ z squared plus open parenthesis the fraction d over 2 close parenthesis squared ] to the power 3 / 2 k ^ .")[$arrow("E") ( z ) = frac(1, 4 π ε_(0)) #h(0.2em) frac(2 q z, attach([ z^(2) + attach(( frac(d, 2) ), t: 2) ], t: 3 "/" 2)) hat("k") .$] + If the source charges are equal and opposite, the vertical components cancel because #math.equation(block: false, alt: "E sub z equals the fraction 1 over 4 π ε sub 0 the fraction q over r squared cos θ minus the fraction 1 over 4 π ε sub 0 the fraction q over r squared cos θ equals 0")[$E_(z) = frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) #h(0.2em) "cos" #h(0.2em) θ − frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) #h(0.2em) "cos" #h(0.2em) θ = 0$] #linebreak() and we get, for the horizontal component of #math.equation(block: false, alt: "E →")[$arrow("E")$], #math.equation(block: true, alt: "E → open parenthesis z close parenthesis, equals the fraction 1 over 4 π ε sub 0 the fraction q over r squared sin θ i ^ minus the fraction 1 over 4 π ε sub 0 the fraction − q over r squared sin θ i ^; equals the fraction 1 over 4 π ε sub 0 the fraction 2 q over r squared sin θ i ^; equals the fraction 1 over 4 π ε sub 0 the fraction 2 q over [ z squared plus open parenthesis the fraction d over 2 close parenthesis squared ] the fraction open parenthesis the fraction d over 2 close parenthesis over [ z squared plus open parenthesis the fraction d over 2 close parenthesis squared ] to the power 1 / 2 i ^ .")[$arrow("E") ( z ) & = frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) #h(0.2em) "sin" #h(0.2em) θ hat("i") − frac(1, 4 π ε_(0)) #h(0.2em) frac("−" q, r^(2)) #h(0.2em) "sin" #h(0.2em) θ hat("i") \ & = frac(1, 4 π ε_(0)) #h(0.2em) frac(2 q, r^(2)) #h(0.2em) "sin" #h(0.2em) θ hat("i") \ & = frac(1, 4 π ε_(0)) #h(0.2em) frac(2 q, [ z^(2) + attach(( frac(d, 2) ), t: 2) ]) #h(0.2em) frac(( frac(d, 2) ), attach([ z^(2) + attach(( frac(d, 2) ), t: 2) ], t: 1 "/" 2)) hat("i") .$] This becomes #math.equation(block: true, alt: "E → open parenthesis z close parenthesis equals the fraction 1 over 4 π ε sub 0 the fraction q d over [ z squared plus open parenthesis the fraction d over 2 close parenthesis squared ] to the power 3 / 2 i ^ .")[$arrow("E") ( z ) = frac(1, 4 π ε_(0)) #h(0.2em) frac(q d, attach([ z^(2) + attach(( frac(d, 2) ), t: 2) ], t: 3 "/" 2)) hat("i") .$] Significance It is a very common and very useful technique in physics to check whether your answer is reasonable by evaluating it at extreme cases. In this example, we should evaluate the field expressions for the cases #math.equation(block: false, alt: "d equals 0")[$d = 0$], #math.equation(block: false, alt: "z ≫ d")[$z ≫ d$], and #math.equation(block: false, alt: "z → ∞")[$z → ∞$], and confirm that the resulting expressions match our physical expectations. Let’s do so: Let’s start with , the field of two identical charges. From far away (i.e., #math.equation(block: false, alt: "z ≫ d close parenthesis ,")[$z ≫ d \) ,$] the two source charges should “merge” and we should then “see” the field of just one charge, of size 2#emph[q]. So, let #math.equation(block: false, alt: "z ≫ d ;")[$z ≫ d ;$] then we can neglect #math.equation(block: false, alt: "d squared")[$d^(2)$] in to obtain #math.equation(block: true, alt: "lim d → 0 E →, equals the fraction 1 over 4 π ε sub 0 the fraction 2 q z over [ z squared ] to the power 3 / 2 k ^; equals the fraction 1 over 4 π ε sub 0 the fraction 2 q z over z cubed k ^; equals the fraction 1 over 4 π ε sub 0 the fraction open parenthesis 2 q close parenthesis over z squared k ^ ,")[$limits("lim")_(d → 0) arrow("E") & = frac(1, 4 π ε_(0)) #h(0.2em) frac(2 q z, attach([ z^(2) ], t: 3 "/" 2)) hat("k") \ & = frac(1, 4 π ε_(0)) #h(0.2em) frac(2 q z, z^(3)) hat("k") \ & = frac(1, 4 π ε_(0)) #h(0.2em) frac(( 2 q ), z^(2)) hat("k") ,$] which is the correct expression for a field at a distance #emph[z] away from a charge 2#emph[q]. Next, we consider the field of equal and opposite charges, . It can be shown (via a Taylor expansion) that for #math.equation(block: false, alt: "d ≪ z ≪ ∞")[$d ≪ z ≪ ∞$], this becomes #math.equation(block: true, alt: "E → open parenthesis z close parenthesis equals the fraction 1 over 4 π ε sub 0 the fraction q d over z cubed i ^ ,")[$arrow("E") ( z ) = frac(1, 4 π ε_(0)) #h(0.2em) frac(q d, z^(3)) hat("i") ,$] which is the field of a dipole, a system that we will study in more detail later. (Note that the units of #math.equation(block: false, alt: "E →")[$arrow("E")$] are still correct in this expression, since the units of #emph[d] in the numerator cancel the unit of the “extra” #emph[z] in the denominator.) If #emph[z] is #emph[very] large #math.equation(block: false, alt: "open parenthesis z → ∞ close parenthesis")[$( z → ∞ )$], then #math.equation(block: false, alt: "E → 0")[$E → 0$], as it should; the two charges “merge” and so cancel out. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ What is the electric field due to a single point particle? #solutionbox[ #math.equation(block: true, alt: "E → equals the fraction 1 over 4 π ε sub 0 the fraction q over r squared r ^")[$arrow("E") = frac(1, 4 π ε_(0)) #h(0.2em) frac(q, r^(2)) hat(r)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Try this #link("https://openstax.org/l/21elefielhocke")[simulation of electric field hockey] to get the charge in the goal by placing other charges on the field. ] === Summary - The electric field is an alteration of space caused by the presence of an electric charge. The electric field mediates the electric force between a source charge and a test charge. - The electric field, like the electric force, obeys the superposition principle - The field is a vector; by definition, it points away from positive charges and toward negative charges. === Conceptual Questions When measuring an electric field, could we use a negative rather than a positive test charge? #solutionbox[ Either sign of the test charge could be used, but the convention is to use a positive test charge. ] During fair weather, the electric field due to the net charge on Earth points downward. Is Earth charged positively or negatively? If the electric field at a point on the line between two charges is zero, what do you know about the charges? #solutionbox[ The charges are of the same sign. ] Two charges lie along the #emph[x]-axis. Is it true that the net electric field always vanishes at some point (other than infinity) along the #emph[x]-axis? === Problems A particle of charge #math.equation(block: false, alt: "2.0 times 10 to the power −8 C")[$2.0 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "C"$] experiences an upward force of magnitude #math.equation(block: false, alt: "4.0 times 10 to the power −6 N")[$4.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "N"$] when it is placed in a particular point in an electric field. (a) What is the electric field at that point? (b) If a charge #math.equation(block: false, alt: "q equals −1.0 times 10 to the power −8 C")[$q = −1.0 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "C"$] is placed there, what is the force on it? #solutionbox[ a. #math.equation(block: false, alt: "E equals 2.0 times 10 squared the fraction N over C")[$E = 2.0 #h(0.2em) × #h(0.2em) 10^(2) frac("N", "C")$] up; #linebreak() b. #math.equation(block: false, alt: "F equals 2.0 times 10 to the power −6 N")[$F = 2.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "N"$] down ] On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 100 N/C. Compare the gravitational and electric forces on a small dust particle of mass #math.equation(block: false, alt: "2.0 times 10 to the power −15 g")[$2.0 #h(0.2em) × #h(0.2em) 10^(−15) #h(0.2em) "g"$] that carries a single electron charge. What is the acceleration (both magnitude and direction) of the dust particle? Consider an electron that is #math.equation(block: false, alt: "10 to the power −10 m")[$10^(−10) #h(0.2em) "m"$] from an alpha particle #math.equation(block: false, alt: "open parenthesis q equals 3.2 times 10 to the power −19 C close parenthesis .")[$( q = 3.2 #h(0.2em) × #h(0.2em) 10^(−19) #h(0.2em) "C" ) .$] (a) What is the electric field due to the alpha particle at the location of the electron? (b) What is the electric field due to the electron at the location of the alpha particle? (c) What is the electric force on the alpha particle? On the electron? #solutionbox[ a. #math.equation(block: false, alt: "E equals 2.88 times 10 to the power 11 N / C")[$E = 2.88 #h(0.2em) × #h(0.2em) 10^(11) #h(0.2em) "N" "/" "C"$]; #linebreak() b. #math.equation(block: false, alt: "E equals 1.44 times 10 to the power 11 N / C")[$E = 1.44 #h(0.2em) × #h(0.2em) 10^(11) #h(0.2em) "N" "/" "C"$]; #linebreak() c. #math.equation(block: false, alt: "F equals 4.61 times 10 to the power −8 N")[$F = 4.61 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "N"$] on alpha particle; #linebreak() #math.equation(block: false, alt: "F equals 4.61 times 10 to the power −8 N")[$F = 4.61 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "N"$] on electron ] Each of the balls shown below carries a charge #emph[q] and has a mass #emph[m]. The length of each thread is #emph[l], and at equilibrium, the balls are separated by an angle #math.equation(block: false, alt: "2 θ")[$2 θ$]. How does #math.equation(block: false, alt: "θ")[$θ$] vary with #emph[q] and #emph[l]? Show that #math.equation(block: false, alt: "θ")[$θ$] satisfies #math.equation(block: true, alt: "sin open parenthesis θ close parenthesis squared tan open parenthesis θ close parenthesis equals the fraction q squared over 16 π ε sub 0 g l squared m")[$"sin" attach(( θ ), t: 2) #h(0.2em) "tan" ( θ ) = frac(q^(2), 16 π ε_(0) g l^(2) m)$]. #figure(figph[Two small balls are attached to threads of length l which are in turn tied to the same point on the ceiling. The threads hang at an angle of theta to either side of the vertical. Each ball has a charge q and mass m.], alt: "Two small balls are attached to threads of length l which are in turn tied to the same point on the ceiling. The threads hang at an angle of theta to either side of the vertical. Each ball has a charge q and mass m.", caption: none) What is the electric field at a point where the force on a #math.equation(block: false, alt: "−2.0 times 10 to the power −6 −C")[$−2.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "−C"$] charge is #math.equation(block: false, alt: "open parenthesis 4.0 i ^ minus 6.0 j ^ close parenthesis times 10 to the power −6 N?")[$( 4.0 hat("i") − 6.0 hat("j") ) #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "N?"$] #solutionbox[ #math.equation(block: true, alt: "E equals open parenthesis −2.0 i ^ plus 3.0 j ^ close parenthesis N/C")[$E = ( −2.0 hat("i") + 3.0 hat("j") ) #h(0.2em) "N/C"$] ] A proton is suspended in the air by an electric field at the surface of Earth. What is the strength of this electric field? The electric field in a particular thundercloud is #math.equation(block: false, alt: "2.0 times 10 to the power 5 N/C .")[$2.0 #h(0.2em) × #h(0.2em) 10^(5) #h(0.2em) "N/C" .$] What is the acceleration of an electron in this field? #solutionbox[ #math.equation(block: true, alt: "F equals 3.204 times 10 to the power −14 N")[$F = 3.204 #h(0.2em) × #h(0.2em) 10^(−14) #h(0.2em) "N"$], #linebreak() #math.equation(block: true, alt: "a equals 3.517 times 10 to the power 16 m / s squared")[$a = 3.517 #h(0.2em) × #h(0.2em) 10^(16) #h(0.2em) "m" "/" "s"^(2)$] ] A small piece of cork whose mass is 2.0 g is given a charge of #math.equation(block: false, alt: "5.0 times 10 to the power −7 C .")[$5.0 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "C" .$] What electric field is needed to place the cork in equilibrium under the combined electric and gravitational forces? If the electric field is #math.equation(block: false, alt: "100 N/C")[$100 #h(0.2em) "N/C"$] at a distance of 50 cm from a point charge #emph[q], what is the value of #emph[q]? #solutionbox[ #math.equation(block: true, alt: "q equals 2.78 times 10 to the power −9 C")[$q = 2.78 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "C"$] ] What is the electric field of a proton at the first Bohr orbit for hydrogen #math.equation(block: false, alt: "open parenthesis r equals 5.29 times 10 to the power −11 m close parenthesis ?")[$( r = 5.29 #h(0.2em) × #h(0.2em) 10^(−11) #h(0.2em) "m" ) ?$] What is the force on the electron in that orbit? (a) What is the electric field of an oxygen nucleus at a point that is #math.equation(block: false, alt: "10 to the power −10 m")[$10^(−10) #h(0.2em) "m"$] from the nucleus? (b) What is the force this electric field exerts on a second oxygen nucleus placed at that point? #solutionbox[ a. #math.equation(block: false, alt: "E equals 1.15 times 10 to the power 12 N / C")[$E = 1.15 #h(0.2em) × #h(0.2em) 10^(12) #h(0.2em) "N" "/" "C"$]; #linebreak() b. #math.equation(block: false, alt: "F equals 1.47 times 10 to the power −6 N")[$F = 1.47 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "N"$] ] Two point charges, #math.equation(block: false, alt: "q sub 1 equals 2.0 times 10 to the power −7 C")[$q_(1) = 2.0 #h(0.2em) × #h(0.2em) 10^(−7) #h(0.2em) "C"$] and #math.equation(block: false, alt: "q sub 2 equals −6.0 times 10 to the power −8 C ,")[$q_(2) = −6.0 #h(0.2em) × #h(0.2em) 10^(−8) #h(0.2em) "C" ,$] are held 25.0 cm apart. (a) What is the electric field at a point 5.0 cm from the negative charge and along the line between the two charges? (b)What is the force on an electron placed at that point? Point charges #math.equation(block: false, alt: "q sub 1 equals 50 μ C")[$q_(1) = 50 #h(0.2em) μ "C"$] and #math.equation(block: false, alt: "q sub 2 equals −25 μ C")[$q_(2) = −25 #h(0.2em) μ "C"$] are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge #math.equation(block: false, alt: "q sub 3 equals 20 μ C")[$q_(3) = 20 #h(0.2em) μ "C"$] situated there? #solutionbox[ If the #math.equation(block: false, alt: "q sub 2")[$q_(2)$] is to the right of #math.equation(block: false, alt: "q sub 1 ,")[$q_(1) ,$] the electric field vector from both charges point to the right. a. #math.equation(block: false, alt: "E equals 2.70 times 10 to the power 6 N / C")[$E = 2.70 #h(0.2em) × #h(0.2em) 10^(6) #h(0.2em) "N" "/" "C"$]; #linebreak() b. #math.equation(block: false, alt: "F equals 54.0 N")[$F = 54.0 #h(0.2em) "N"$] ] Can you arrange the two point charges #math.equation(block: false, alt: "q sub 1 equals −2.0 times 10 to the power −6 C")[$q_(1) = −2.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C"$] and #math.equation(block: false, alt: "q sub 2 equals 4.0 times 10 to the power −6 C")[$q_(2) = 4.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C"$] along the #emph[x]-axis so that #math.equation(block: false, alt: "E equals 0")[$E = 0$] at the origin? Point charges #math.equation(block: false, alt: "q sub 1 equals q sub 2 equals 4.0 times 10 to the power −6 C")[$q_(1) = q_(2) = 4.0 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C"$] are fixed on the #emph[x]-axis at #math.equation(block: false, alt: "x equals −3.0 m")[$x = −3.0 #h(0.2em) "m"$] and #math.equation(block: false, alt: "x equals 3.0 m .")[$x = 3.0 #h(0.2em) "m" .$] What charge #emph[q] must be placed at the origin so that the electric field vanishes at #math.equation(block: false, alt: "x equals 0 , y equals 3.0 m ?")[$x = 0 , y = 3.0 #h(0.2em) "m" ?$] #solutionbox[ There is #math.equation(block: false, alt: "45 °")[$45 "°"$] right triangle geometry. The #emph[x]-components of the electric field at #math.equation(block: false, alt: "y equals 3 m")[$y = 3 #h(0.2em) "m"$] cancel. The #emph[y]-components give #math.equation(block: false, alt: "E open parenthesis y equals 3 m close parenthesis equals 2.83 times 10 cubed N / C")[$E ( y = 3 #h(0.2em) "m" ) = 2.83 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "N" "/" "C"$]. #linebreak() At the origin we have a a negative charge of magnitide #linebreak() #math.equation(block: false, alt: "q equals −2.83 times 10 to the power −6 C")[$q = −2.83 #h(0.2em) × #h(0.2em) 10^(−6) #h(0.2em) "C"$]. ]