Chapter 2. Coulomb s Law

Transcription

1 Chapter. Coulomb s Law PHY 4, Fall 7 Coulomb' s Law for point charges. Coulomb' s Law, coordinate-free Coulomb s Law is usuall stated first without reference to coordinate sstem: The force (vector) on a point charge Q, caused b point charge q, is magnitude: F Q = q Q d direction: along the line joining the charges, attractive for unlike charges, repulsive for like charges, where d is the distance between the charges q and Q. We consider q to be the source charge, the cause of the electrical force on the test charge Q. The equation is smmetric with respect to q and Q, so we could equall write F q = q Q, where d Q is considered the source charge and q, now considered the test charge, feels the force produced b Q. This smmetr is just an expression of Newton s third law.. Coulomb' s Law, with a coordinate sstem Practical calculations almost alwas require a coordinate sstem, so let s translate the description of the electrical force given in Section into mathematics in a Cartesian coordinate sstem. The force on point charge Q at point r = (x,, z) caused b a single point charge q at point r = (x,, z ) is given b F Q (r) = Q q 'r( r ) = Q q (r- r ) r- r ' 3 where r = vector from origin to test charge Q: r = r r ) = x x ) + ) + z z ). r = vector from origin to source charge q: r = r r ) = x x ) + ) + z z ) r = r - r = vector from source charge to test charge: r r r x x ) ) z z ) x x ) ) z z )

2 r r r = vector from source charge to test charge: r = r - r = x x ) + ) + z z ) - x x ) + ) + z z ) Note that the magnitude of the script-r vector, 'r(, is the distance between the charges, called d in the coordinate-free version. You might be thinking Wh get so complicated? Just put the source charge at the origin?. In that case r = and r = r - r = r r ), which gives F Q (r) = Q q r r ) (source charge q at the origin). Note that the vector r = r r ) is the radial position vector, from the origin to a point (x,, z) a distance r = x + + z awa. The unit vector r ) is in the radial direction, but does not have a unique direction like the Cartesian unit vectors x ), ), and z ) whose directions are fixed, i.e. it points from the origin to (an) point a distance r awa. points: Example: Tr plotting (or visualizing in our brain) the vector r = r r ) for the following a. r = (x,, z) = (,, ) b. r = (x,, z) = (,, ) c. r = (x,, z) = (,, ) Example: With q at (, 3, ) and Q at (5/,, 5). Find F Q q. Answer: F Q q = q Q 8 π ϵ / (-, 6, - 6).

3 3 3. Superposition With multiple charges, the pairwise force adds vectoriall -- this is known as the principle of superposition, and it makes electrostatics relativel simple. Here s an example for source charges: Let q and q be considered the sources and Q be the test charge, then the principle of superposition sas that the total force on Q is the vector sum of the individual forces caused b each source charge: F Q = F Q + F Q = q Q r Q r ) Q + q Q r Q r ) Q, where r Q is the distance between q and Q, and r Q is the distance between q and Q. Each force term is just Coulomb s law written for the respective source charge. For N source charges q i (with i =,, 3,..., N), the total force on test charge Q caused b all the source charges q i at positions r i is F Q (r) = Q N 4 π ϵ i= q i r r# i = Q N q i (r- r i ) i 4 π ϵ i= r- r i Coulomb s Law for N source charges ' 3 Here the unit vectors r ) i Q are in the direction along the line joining source charge q i and test charge Q as described above. Note that this is a vector sum! Not understanding this is a significant source of errors in PHY 4! 3. Example: source charges, solved with intuition & smmetr Here s an example for source charges: Let q be at r = (,, ) and q = q at r = (-,, ) be considered the source charges exerting a force on the test charge Q, located r = (,, ). If it helps, draw ourself a diagram:

4 4 Strateg : Intuition, smmetr, geometr The individual forces on Q, F and F, are shown in green at the top of the diagram. If ou notice the smmetr in this problem, ou can use it to save ou work: since the source charges are equal ans located smmetricall in x relative to Q, it s prett clear that the x-components of the forces cancel! Therefore, the total force on Q is in the -direction. Moreover, look at the geometr of these force vectors for a minute and ou might be able to see that the -components aren t that hard to get. From the F and F vectors and simple geometr, ou get F = F cosα and F = +F cosα, so F Q = F cosα + F cosα Again we appeal to smmetr to sa that F = F and α = α, giving F Q = F cosα. The force magnitudes are given b Coulomb s law: F = q Q r = q Q (x ) + and the angles b cosα = (x ) + / (look at the lower triangles). Putting this all together will ield F Q = F cosα = q Q 4 = q Q, π ϵ (x ) + (x ) + π ϵ (x ) + 3 so the final result is F Q = π ϵ q Q (x ) + 3 ), or, with the given numbers, x =, x = -, = plugged in: F Q = π ϵ q Q 3 ). End of Class 3, Fall 7 3. Example: source charges, solved with the mathematical framework Strateg : The mathematical framework. This approach alwas works if ou appl it correctl, but it can also be the most tedious to calculate. For this problem, the phsics is prett straightforward: the total force on Q is the

5 5 calculate. For this problem, the phsics is prett straightforward: the total force on is the superposition of the forces due to each source charge. The mathematical framework is the force superposition equation, above, which looks like F Q = F Q + F Q = q Q r r ) + q Q r r ). Using the given information, the expressions for the positions and the distances between charges become r = vector from origin to test charge Q: r = ), r = vector from origin to source charge q : r = x x ), r = vector from origin to source charge q : r = x x ). From these, we need to calculate the script-r separation vectors, r = r - r = vector from source charge to test charge: r = r - r = ) - x x ) r = r - r = vector from source charge to test charge: r = r - r = ) - x x ). For the superposition equation, ou need the magnitudes of these and their unit vectors. In Cartesian coordinates, the magnitudes are r = (x ) +, and r = (x ) +. B definition, the unit vectors are of magnitude with direction the same as the vectors, so ou can divide each vector b its magnitude: r ) = r = r ) - x x ), and r ) = r = ' (x ) + r ) - x x ), ' (x ) + and a little algebra ields the individual forces, F = q Q r r ) = q Q x + 3 ) - x x ), and F = q Q r r ) = q Q x + ) - x 3 x ), and the total force on Q is the sum:

6 6 F = F + F = q Q x x + 3 ) - q Q x x x x + 3 x ). This is a more general solution than we need for this problem: we haven t et used the fact that x = - x and q = q. Plugging this given data in to the framed equation above gives F = F + F = q Q x x + 3 ) - q Q - x x x x + 3 x ). The second term vanishes (thus proving the smmetr used in the above method) and the first term leaves F = π ϵ q Q x + 3 ) This is, of course, the same as the intuitive strateg gave. Which strateg is best? The intuitive calculation took 3 lines (plus a little hidden algebra), while the math framework strateg took much more, as is almost alwas true. But the intuitive strateg ma not be so intuitive to ou, while the framework usuall suggests a solution algorithm and doesn t depend on having a strong phsical intuition. In fact, one wa to help build our intuition is to do problems (using the framework) and notice the smmetries and simplifications that appear naturall. 4. Aside : Visualizing vectors with diagrams There are two figures above, used to help ou visualize vectors in D and 3D space. You can draw such diagrams b hand, for use in solving homework or exam problems, and if ou re careful enough that should suffice. If ou re interested in more precise and estheticall pleasing diagrams, ou ll need a software tool. I used Mathematica to draw the diagrams above. The main commands are: Graphics or Graphics3D: to draw the diagram Line, Arrow, Point, Circle: to define the components of the diagram Text: to label parts of the diagram Legended, LineLegend, etc: to create a legend for our diagram Here are a couple examples.

7 7 4. Code for D diagram in Section 3. above Here s some sample code for producing the D vector diagram, used for the example in Section 4 above: In[]:= (* Define basic quantities: * ) origin = {, }; source = {, }; source = {-, }; field = {, }; x = source[[]]; = source[[]]; x = source[[]]; = source[[]]; x = field[[]]; = field[[]]; d =.5; scriptr = x + ; scriptr = x + ; Fx = - d x / scriptr; F = + d / scriptr; Fx = - d x / scriptr; F = + d / scriptr; (* Make the diagram, one component at a time: * ) Graphics {PointSize[.5], Point[{source, source, field}]}, (* Draw charges and origin * ) {Circle[origin,.4]}, {Red, Thick, Arrow[{origin, source}]}, (* Draw source and test charge position vectors * ) {Red, Thick, Arrow[{origin, source}]}, {Blue, Thick, Arrow[{origin, field}]}, {Black, Thickness[.3], Arrow[{source, field}]}, (* Draw difference vectors, "script- r" * ) {Black, Thickness[.3], Arrow[{source, field}]}, {Green, Thickness[.5], Arrow[{field, {Fx, F}}]}, (* Draw force vectors * ) {Green, Thickness[.5], Arrow[{field, {Fx, F}}]}, {Text["Source charge q ", {x +.5, +.}]}, (* Label charges, forces, etc. * ) {Text["Source charge q ", {x -.4, +.}]}, {Text["Test charge Q", field, {-.5, }]}, {Text["Origin", origin + {.,.}]}, Text "F ", {x + d -., + d -.5},

8 8 Text "F ", {x - d +., + d -.5}, {Text["α ", {x -., -.3}]}, {Text["α ", {x +., -.3}]}, (* Adjust plot parameters for tweaking the look of the diagram: * ) PlotRange {{-.6,.6}, {,.6}}, Axes True, AxesLabel {"x", ""}, AxesStle Thickness[.], TicksStle Directive[FontOpacit, FontSize ] F.5 F. Test charge Q Out[]= αα αα.5 Source charge q Source charge q Origin x Ver similar code can be used to draw the electric field diagram in Section 5. below. 4. Code for 3D diagram in Section above Here s some sample code for producing the 3D vector diagram, used in Section. This is a simpler sstem (onl one source charge), but the 3D plot adds a little complexit.

9 9 In[]:= (* Define basic quantities: * ) origin = {,, }; source = {, 3, }; field = {5 /, / 3, 5}; projectionq = source - {,, source[[3]]}; projectionq = field - {,, field[[3]]}; (* Make the diagram, one component at a time: * ) Legended Graphics3D[ {{PointSize[Large], Point[{source, field}]}, (* Draw charges * ) {Red, Thick, Arrow[{origin, source}]}, (* Draw charge position vectors * ) {Blue, Thick, Arrow[{origin, field}]}, {Black, Thickness[.4], Arrow[{source, field}]}, (* Draw difference vector, "script- r" * ) {PointSize[Medium], Gra, Point[{projectionq, projectionq}]}, {Dashed, Gra, Line[{source, projectionq}]}, (* Draw dashed projection lines * ) {Dashed, Gra, Line[{origin, projectionq}]}, (* used to help see the location of * ) {Dashed, Gra, Line[{field, projectionq}]}, (* the vectors in 3D space * ) {Dashed, Gra, Line[{origin, projectionq}]}, {Text["Source charge, q", source, {-.3,.5}]}, (* Label charges * ) {Text["Field point, P", field, {-.3, }]}}, (* Adjust plot parameters for tweaking the look of the diagram: * ) PlotRange {{, 4}, {, 4}, {, 5}}, Axes True, AxesEdge {{-, - }, {-, - }, {-, - }}, AxesLabel {"x", "", "z"}, AxesStle Thickness[.3], ViewPoint {6,,.5}], (* Make the legend: * )LineLegend {Red, Blue, Black}, "Source vector, r ", "Field vector, r ", "'script- r' vector = r - r "

10 Field point, P 4 z Source vector, r Out[]= Field vector, r Source charge, q 'script- r' vector = r - r x Electric Field You ma have noticed that the test charge is kind of irrelevant in these problems, it s just a multiplicative factor in ever equation since the force is proportional to Q. Wh not eliminate it b dividing it out? You can easil define the force per unit charge: E(r) = F Q(r) Q, or, more rigorousl, we take the limit as the test charge Q shrinks to zero: E(r) = lim Q F Q(r) Q So that the superposition version of Coulomb s law becomes: E (r) = N i= q i r i ) r i = N q i (r- r i ) i= (Coulomb s Law for Electric field E) r- r i ' 3 A common wa of thinking of it is that E is the limiting case of F Q Q as the test charge Q is

11 shrunk to zero, leaving onl a force field at position r, which would produce a force F Q = Q E on a test charge Q, if ou put one there. This electric field E is produced b all the source charges, as Coulomb s Law superposition principle states. 5. Two source charge example above, but for E We consider the same two source charges: Let q be at r = (,, ) and q = q at r = (-,, ), but now ou can think of them as producing an electric field E at the point r = (,, ). I ll use the notation P for the field point, and re-draw the diagram. As ou can see, it s essentiall identical to the force on test charge example above, with Q replaced b P and forces F & F replaced b fields E & E : The solution is also identical, given these replacements, so the answer will also be the same, divided b Q: E = lim Q F Q = π ϵ q x + 3 ) If ou need to know the force on some test charge Q it would be F Q = Q E, while the force on a different test charge Q would be F Q = Q E. Knowing the field produced b the source charges therefore allows ou to calculate the force on an test charge located at P. (Code for the diagram is hidden) 5. Example: A different problem with source charges Let s put the one source charge on the x-axis and one on the -axis. Let q be at r = (x,, ) and q at r = (,, ) be considered the source charges producing a field E to be measured along the z-axis at r = (,, z). Here s a 3D diagram:

12 along the z-axis at. Here s a 3D diagram: The smmetr of the previous example is no longer valid here, there will be components is all three Cartesian directions. With experience, ou can use the intuition/geometr method, but let s start with the math framework approach. Begin with the basic position vectors: r = vector from origin to field point P: r = z z ) r i = vector from origin to source charge q i : r = x x ), r = x ) r i = r - r i = vector from q i to P: r = r - r = z z ) - x x ), r = r - r = z z ) - ) 'r i ( = 'r - r i ( = distance between q i and Q: ' r ( = z + (x ) /, ' r ( = z + ( ) / The method tells us that the field is then given b E (r) = N i= so we just handle each term separatel: q i r i ) r i = N q i (r- r i ) i=, r- r i ' 3 E(r) = q ) q r r ) q + r r = r 4 + q r π ϵ z +(x ) z +(x ) / z +( ) z +( ) / = or, combining like components, q z +(x ) 3/ z z) - x x ) + q z +( ) 3/ z z) - ) E (r) = q z +(x ) 3/ + q z +( ) 3/ z z) - q z +(x ) 3/ x x ) - q z +( ) 3/ )

13 3 Just for fun, let s plot the components of this field as a function the location of the field point on the z-axis. Taking numbers used for the diagram above, x =, =, and setting q = q for simplicit, I put the solution for E(z) into Mathematica language: In[3]:= Ezqx = - xp z + xp 3/ ; Ezq = - p z + p 3/ ; Ezqz = - z Ezqx - z xp p Ezq; Row "E (z) = ", Ez = {Ezqx, Ezq, Ezqz} /. {xp, p } Plot Ez, {z,, 5}, Frame True, FrameLabel "z", "E (z)", FrameStle Thickness[.4], PlotLabel "q = q ", AxesStle Thickness[.4], PlotLegends LineLegend Automatic, "E x ", "E ", "E z " Out[4]= E(z) = -, -, + z 3/ 4 + z 3/ z + z + z 3/ 4 + z 3/.5 q = q. Out[5]= E (z) E x E -.5 E z z I've actuall plotted the quantit E / q, where q = q = q, to avoid dealing with potentiall large or small values. If ou need to know the actual field (in standard units of N/C, e.g.), ou ll need to multipl the plot values accordingl. You can see from the graph that the z-component of E is positive and the other two components are negative, all apparentl going to zero as z. Build our phsical intuition: Can ou explain wh these observed behaviors should be true? Now, consider opposite charges q = - q. Does that change anthing? Let s see:

14 4 In[6]:= Ezqx = - Ezq = - xp z + xp 3/ ; - p z + p 3/ ; Ezqz = - z Ezqx - z xp p Ezq; Row "E (z) = ", Ez = {Ezqx, Ezq, Ezqz} /. {xp, p } Plot Ez, {z,, 5}, PlotRange All, Frame True, FrameLabel "z", "E (z)", FrameStle Thickness[.4], AxesStle Thickness[.4], PlotLabel "q = - q ", PlotLegends LineLegend Automatic, "E x ", "E ", "E z " Out[7]= E(z) = -, + z 3/ 4 + z 3/, z z - + z 3/ 4 + z 3/ q = - q.. -. E x Out[8]= E (z) -.4 E -.6 E z z It looks like E has changed sign (make sense?), and the magnitude of E z has decreased (make sense?). (Code for Section 5. vector diagram is hidden) End of Class 4, Fall 7 6. Aside : Visualizing the electric field Take a charge q at the origin. It s E field is given b Coulomb s Law, E(r) = q r r ), where r = r r ) is the radial position vector to the field point P: (x,, z). Note that r is also the

15 5 spherical radial coordinate. The magnitude r = x + + z and the unit vector r ) is defined as the full vector over its magnitude, r ) = r / r = x x ) + ) + z z ) r. Therefore, E(r) = q x + +z 3/ x x) + ) + z z ) I want to plot the field onl in the x-plane, so to slice it in the x-plane, we set z = : E (r) = q x + 3/ x x) + ). This is the vector field I will plot. 6. Using sample vectors to visualize E(r) Rewriting the field in Mathematica code, In[33]:= Out[35]= Clear[x, ]; q = ; Er = q x + 3/ ; rhat = {x, }; Evec = Simplif[Er rhat /. q q] x, x + 3/ x + 3/ There are several was to visualize a vector field such as E. One wa is to draw vectors representing the field at various sample points in space: a vector field plot. Mathematica does this as follows: In[36]:= VectorPlot[Evec, {x, - 3, 3}, {, - 3, 3}, ImageSize Small] 3 Out[36]= This is not useful. What s happening is the field magnitude is huge near the charge ( ) and decreases so quickl that the plot is dominated b one vector ver close to the charge. There are options that ou might be able to use to get a meaningful vector plot, but I find the plot less useful than field line plots (see below), so I m going to jump to them. If ou re interested, check out the man options on the VectorPlot command that allow ou to refine this form of visualization. For example here s one suggestion from the online help:

16 6 In[37]:= VectorPlot[Evec, {x, - 3, 3}, {, - 3, 3}, VectorScale {Automatic, Automatic, #}] & {Function[If[#5 >, None, #5]]} 3 Out[37]= Using field lines to visualize E(r) from a single source charge Field lines are curves that are locall parallel to the electric field, giving a nice visualization of the direction of the field. In a standard field line plot, the densit of lines, (e.g. the number of lines passing through a unit area in 3-D or unit length in -D) is proportional to the magnitude of the field. Mathematica has a command to plot such curves, StreamPlot, but it doesn t necessaril plot the densit-magnitude relation as we d like. Nonetheless, such plots can help us visualize the field. The command is named after velocit streamlines fluid dnamics, which are locall parallel to the direction of the velocit field of a fluid. To plot our single charge field lines, just replace the plot command with StreamPlot:

17 7 In[38]:= StreamPlot[Evec, {x, - 3, 3}, {, - 3, 3}, FrameLabel {"x", ""}, PlotLabel "Positive Point Charge", Epilog {PointSize[Large], Green, Point[{, }]}] Positive Point Charge 3 Out[38]= x With StreamPlot, the arrows line up to form electric field lines, S described above. To help visualize, I added a green dot where the charge is located. The direction of flow of the E field is now eas to see, but Mathematica s default is to tr to keep the densit of lines approximatel uniform across the plot (apparentl that s the standard for fluid streamlines). This allows the following interpretation errors:. The densit of lines is not proportional to the field magnitude. Streamlines just end or begin in empt space The densit of lines depicting the field magnitude is useful, as described above. Point here is more of a problem: b definition, lines of E should start on positive charges (or at infinit) and end on negative charges (or at infinit). In fact, if this fails, Point will also fail. Without further work, we ll just have to live with these problems, being careful not to misinterpret the plot. The plot above is for a positive charge. A negative charge looks like this:

18 8 In[39]:= StreamPlot[- Evec, {x, - 3, 3}, {, - 3, 3}, FrameLabel {"x", ""}, PlotLabel "Negative Point Charge", Epilog {PointSize[Large], Magenta, Point[{, }]}] Negative Point Charge 3 Out[39]= x The obvious difference is that E flows awa from a positive charge and toward a negative charge 6.3 Using field lines to visualize E(r) from source charges You can use the same method to plot field lines for the field produced b two point charges, sa of opposite signs, with the positive charge at the origin and the negative charge shifted along the x-axis b an amount x shift, follows:

19 9 In[4]:= q = - ; xshift = ; Row[{"E (* Choose sign and x- shif for the second charge * ) = ", Evec = Simplif[Er rhat /. {q q, x (x - xshift)}]}] StreamPlot[Evec, {x, -.5, 4.5}, {, - 3, 3}, FrameLabel {"x", ""}, Epilog {PointSize[Large], Magenta, Point[{xshift, }]}, ImageSize Small] - + x Out[4]= E = -, - (- + x) + 3/ (- + x) + 3/ 3 Out[4]= x OK, the charge is shifted right b units. The total field is just the sum of the individual fields, b superposition:

20 In[43]:= xbdr = ; xmin = - xbdr; xmax = xshift + xbdr; min = - xbdr; max = xbdr; StreamPlot[Evec + Evec, {x, xmin, xmax}, {, min, max}, FrameLabel {"x", ""}, PlotLabel "E of oppositel charged point charges", AspectRatio (max - min) / (xmax - xmin), Epilog {{PointSize[Large], Green, Point[{, }]}, {PointSize[Large], Magenta, Point[{xshift, }]}}, ImageSize Large] E of oppositel charged point charges Out[43]= I hope our phsical intuition agrees with this plot: near the charges the lines of E look approximatel radial, but further out, the bend so the lines that began on the positive charge, end on negative charge. Some of them have to go a long wa awa from the charges, but the will come back (the onl exception being the line that goes out from +q along the - x axis: it goes to infinit and the line that comes in to the negative along the +x axis, began at infinit. x B the wa, this pair of oppositel charged points is called a dipole. You can use a Talor series to show that the field of a dipole decreases like r 3 for large distances r awa from the dipole. We ll see more on dipoles in Chapter 3. You now have a visualization tool ou can use to plot an number of point charges (as long as ou can calculate them mathematicall!). For example, make the negative point charge 5 times the charge as the positive point charge:

21 You now have a visualization tool ou can use to plot an number of point charges (as long as ou can calculate them mathematicall!). For example, make the negative point charge 5 times the charge as the positive point charge: In[44]:= xbdr = 3; xmin = - xbdr; xmax = xshift + xbdr; min = - xbdr; max = xbdr; StreamPlot[Evec + 5 Evec, {x, xmin, xmax}, {, min, max}, FrameLabel {"x", ""}, PlotLabel "E of oppositel charged point charges, q = - 5q ", AspectRatio (max - min) / (xmax - xmin), Epilog {{PointSize[Large], Green, Point[{, }]}, {PointSize[Large], Magenta, Point[{xshift, }]}}, ImageSize Large] E of oppositel charged point charges, q = - 5q 3 Out[44]= The plot is noticeabl different: now positive charge field lines turn around fairl near the +q charge, and head straight for the negative charge. There is also now a stagnation point (where a fluid would have zero velocit) around the point (-, ). The field isn t flowing but imagining it as analogous to a fluid is not a bad intuition to determine qualitative behavior. The E field stagnates there, too, i.e. it vanishes at that point. Near the charges the lines of E look approximatel radial, but further out, the bend so the lines that began on the positive charge, turn around quickl and end on negative charge. x

22 6.4 Aside: Fixing the line densit problem b specifing each streamline StreamPlot s failure to exhibit the densit of lines proportional to the magnitude of the field is due to it choosing streamlines such that the densit of line is roughl uniform. However, the StreamPoints option allows ou to specif where to draw streamlines. The point charge field is smmetric enough that we can get closer to the correct densitmagnitude relation b specifing that each streamline pass through a point on a fixed circle centered at the charge and separated b a fixed azimuthal angle d ϕ. Below is the code to do this for points on a circle of radius rp and angular separation dϕ. The arra sptsϕ then holds the points, which are used in the StreamPoints sptsϕ option on the StreamPlot command: In[45]:= (* Define streamline points: equall spaced in ϕ around a circle of radius rp; dϕ = ϕ interval for drawing streamlines * ) dϕ = π / 6; rp =.5; sptsϕ = Table[rp {Cos[ϕ], Sin[ϕ]}, {ϕ,, π - dϕ, dϕ}]; StreamPlot[Evec, {x, - 3, 3}, {, - 3, 3}, FrameLabel {"x", ""}, StreamPoints sptsϕ, Epilog {{PointSize[Large], Green, Point[{, }]}, {Red, Point[sptsϕ]}}] 3 Out[46]= x I drew red dots at the points I defined for each streamline. I think ou ll agree that this plot makes the field more intuitive: no field lines disappearing or magicall appearing; and the densit of lines is now at least consistent with the field magnitude (e.g. greater near the charge than far awa). Can ou use the same method to get meaningful field lines for the two-charge case? It s less obvious since the radial smmetr is no longer present except ver near the charges,

23 3 less obvious since the radial smmetr is no longer present except ver near the charges, and it turns out that this method woks less well ver close (for other reasons, which require further options to be chosen - more complicated than I want to get into here). Let s tr a compromise: chose streamlines close, but not too close. For the streamlines chosen for the single charge plot, we get the following, In[47]:= xbdr = ; xmin = - xbdr; xmax = xshift + xbdr; min = - xbdr; max = xbdr; StreamPlot[Evec + Evec, {x, xmin, xmax}, {, min, max}, FrameLabel {"x", ""}, PlotLabel " Point Charges, oppositel charged", StreamPoints sptsϕ, AspectRatio (max - min) / (xmax - xmin), Epilog {{PointSize[Large], Green, Point[{, }]}, {PointSize[Large], Magenta, Point[{xshift, }]}, {Red, Point[sptsϕ]}}, ImageSize Medium] Point Charges, oppositel charged Out[47]= x This looks prett good near the positive charge, but lines are missing near the negative charge, so let s add some points that determine field lines there: In[48]:= sptsϕ = Table[rp {Cos[ϕ], Sin[ϕ]} + {xshift, }, {ϕ,, π - dϕ, dϕ}];

24 4 In[49]:= xbdr =.5; xmin = - xbdr; xmax = xshift + xbdr; min = - xbdr; max = xbdr; sptstot = Join[sptsϕ, sptsϕ]; StreamPlot[Evec + Evec, {x, xmin, xmax}, {, min, max}, FrameLabel {"x", ""}, PlotLabel "E of oppositel charged point charges", StreamPoints sptstot, AspectRatio (max - min) / (xmax - xmin), Epilog {{PointSize[Large], Green, Point[{, }]}, {PointSize[Large], Magenta, Point[{xshift, }]}, {Red, Point[sptstot]}}, ImageSize Large] E of oppositel charged point charges Out[5]= This looks prett good. The streamline points are defined close enough to the charges that the field is still approximatel radial. Even though there aren t man field lines, ou can get a good idea of what the field looks like. Explore more: what if the charges are the same sign? x

25 5 In[5]:= xbdr =.5; xmin = - xbdr; xmax = xshift + xbdr; min = - xbdr; max = xbdr; sptstot = Join[sptsϕ, sptsϕ]; StreamPlot[Evec + (- Evec), {x, xmin, xmax}, {, min, max}, FrameLabel {"x", ""}, PlotLabel "E of like charged point charges", StreamPoints sptstot, AspectRatio (max - min) / (xmax - xmin), Epilog {{PointSize[Large], Green, Point[{, }]}, {PointSize[Large], Magenta, Point[{xshift, }]}, {Red, Point[sptstot]}}, ImageSize Large] E of like charged point charges Out[5]= Looks good. Note the stagnation point between the charges with nearb field lines deflected toward ±. The technique seems robust enough to explore a bit further. Tr like charges with q = 5 q : x

26 6 In[53]:= xbdr =.5; xmin = - xbdr; xmax = xshift + xbdr; min = - xbdr; max = xbdr; sptstot = Join[sptsϕ, sptsϕ]; StreamPlot[Evec + (- 5 Evec), {x, xmin, xmax}, {, min, max}, FrameLabel {"x", ""}, PlotLabel "E of like charged point charges, q = 5q ", StreamPoints sptstot, AspectRatio (max - min) / (xmax - xmin), Epilog {{PointSize[Large], Green, Point[{, }]}, {PointSize[Large], Magenta, Point[{xshift, }]}, {Red, Point[sptstot]}}, ImageSize Large] E of like charged point charges, q = 5q Out[54]= Check our electric field intuition: what do ou see here? x