Ionic Potentials across Cell Membranes Conceptual Question

Transcription

1 PH213 Chapter 29 Solutions Ionic Potentials across Cell Membranes Conceptual Question Description: Short conceptual problem dealing with ionic potentials across cell membranes In its resting state, the membrane surrounding a neuron is permeable to potassium ions but only slightly permeable to sodium ions Thus, positive K ions can flow through the membrane in an attempt to equalize K concentration, but Na ions cannot as quickly This leads to an excess of Na ions outside of the cell If the space outside the cell is defined as zero electric potential, then the electric potential of the interior of the cell is negative This resting potential is typically about 80 A schematic of this situation is shown in the figure In response to an electrical stimulus, certain channels in the membrane can become permeable to Na ions Due to the concentration gradient, Na ions rush into the cell and the interior of the cell reaches an electric potential of about 40 This process is termed depolarization In response to depolarization, the membrane again becomes less permeable to Na ions, and the K ions flow out of the interior of the cell through channels established by the positive electric potential inside of the cell This then reestablishing the resting potential This is termed repolarization Only a small percentage of the available Na and K ions participate in each depolarization/repolarization cycle, so the cell can respond to many stimuli in succession without depleting its "stock" of available Na and K ions A graph of an electric potential inside a cell vs time is shown in the next figure

2 depolarization/repolarization cycle Part A for a single During the resting phase, what is the electric potential energy of a typical Na ion outside of the cell? A1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, )

3 Part B During the resting phase, what is the electrical potential energy of a typical K ion inside of the cell? B1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, ) 40

4 0 Part C During depolarization, what is the work done (by the electric field) on the first few Na ions that enter the cell? C1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, ) C2 Algebraic sign of the work In general, work is defined as the product of the force applied parallel (or antiparallel) to the displacement of an object Thus, The work done by a force is positive if the force and the displacement are parallel; it is negative if the force and displacement are opposite in direction C3 Magnitude of the work Work transfers energy into or out of a system Therefore, in the absence of other energy transfers, the magnitude of the work done on an object is equal to the magnitude of the object s change in energy Since the primary form of energy present in this example is electric potential energy, the magnitude of the work done is equal to the change in the ion s electric potential energy

5 Part D During repolarization, what is the work done (by the electric field) on the first few K ions that exit the cell? D1 The electron volt Electric potential energy is defined as The electric charge on individual particles is always a multiple of the fundamental charge charge on a single proton) Rather than substituting a numerical value for, it is often more convenient to use the constant as a unit Thus, a proton located at a potential of 100 has energy (the which can be written as, or Thus, the proton has 100 electron volts of energy (Electron volts can be converted to the more traditional unit of energy, the joule, by multiplying by the conversion factor and recalling that Thus, ) D2 Algebraic sign of the work

6 In general, work is defined as the product of the force applied parallel (or antiparallel) to the displacement of an object Thus, The work done by a force is positive if the force and the displacement are parallel; the work done is negative if the force and displacement are opposite in direction D3 Magnitude of the work Work transfers energy into or out of a system Therefore, in the absence of other energy transfers, the magnitude of the work done on an object is equal to the magnitude of the object s change in energy Since the primary form of energy present in this example is electric potential energy, the magnitude of the work done is equal to the change in the ion s electric potential energy Electric Potential Ranking Task Description: Short conceptual problem involving electrical potentials of point charges (ranking task)

7 In the figure there are two point charges, and There are also six positions, labeled A through F, at various distances from the two point charges You will be asked about the electric potential at the different points (A through F) Part A Rank the locations A to F on the basis of the electric potential at each point Rank positive electric potentials as higher than negative electric potentials A1 Definition of electric potential The electric potential surrounding a point charge is defined by where is the source charge creating the electric potential and is the distance between the source charge and the point of interest If more than one source is present, determine the electric potential from each source and sum the results, A2 Conceptualizing electric potential Because positive charges create positive electric potentials in their vicinity and negative charges create negative potentials in their vicinity, electric potential is sometimes visualized as a sort of "elevation" Positive charges represent mountain peaks and negative charges deep valleys In this picture, when you are close to a positive charge, you are "high up" and have a higher positive potential Conversely, near a negative charge, you are deep in a "valley" and have a negative potential The utility of this picture becomes clearer when we begin to think of charges moving through a region of space containing an electric potential Just as particles naturally roll downhill, converting gravitational potential energy into kinetic energy, positively charged particles naturally "roll downhill" as well, toward regions of lower electric potential, converting electrical potential

8 "roll downhill" as well, toward regions of lower electric potential, converting electrical potential energy into kinetic energy Rank the locations from highest to lowest potential To rank items as equivalent, overlap them View Change in Electric Potential Ranking Task Description: Short conceptual problem related to the electric potential difference between pairs of points (ranking task) In the diagram below, there are two charges of and and six points (a through f) at various distances from the two charges rank changes in the electric potential along paths between pairs of points Part A You will be asked to Using the diagram to the left, rank each of the given paths on the basis of the change in electric potential Rank the largest-magnitude positive change (increase in electric potential) as largest and the largest-magnitude negative change (decrease in electric potential) as smallest A1 Change in electric potential Determining the change in electric potential along some path involves determining the electric potential at the two end points of the path, and subtracting:

9 potential at the two end points of the path, and subtracting: A2 Determine the algebraic sign of the change in potential The path from point d to point a results in a positive change in electric potential Which of the other paths also involves a positive change in electric potential (ie, electric potential that increases along the path)? from b to a from f to e from c to d from c to e from c to b A3 Conceptualizing changes in electric potential Since positive charges create large positive electric potentials in their vicinity and negative charges create negative potentials in their vicinity, electric potential is sometimes visualized as a sort of elevation Positive charges represent mountain peaks and negative charges deep valleys In this picture, when you are close to a positive charge, you are high up and have a large positive potential Conversely, near a negative charge you are deep in a valley and have a negative potential Thus, changes in electric potential can be thought of as changes in elevation The change is positive if you are moving uphill and the change is negative if you move downhill The farther you travel either uphill or downhill, the larger the magnitude of the change in electric potential Rank from largest to smallest To rank items as equivalent, overlap them View Not So Fast! Description: A charge is moving in the electric field of other point charges; use energy considerations to find its initial velocity Four point charges, fixed in place, form a square with side length

10 Part A The particle with charge is now released and given a quick push; as a result, it acquires speed Eventually, this particle ends up at the center of the original square and is momentarily at rest If the mass of this particle is, what was its initial speed? A1 How to approach the problem Use the law of conservation of energy for the particle with charge A2 Finding the potential energy Find the potential energy of the particle with charge due to each of the other three charged particles; then use the principle of superposition You have to complete the entire procedure twice, of course: for the initial and the final moments A3 Find the initial potential energy Find the initial potential energy of the particle with charge A31 Formula for the electric potential energy

11 The formula for the electric potential energy between charges and separated by distance is, where Express your answer in terms of,, and appropriate constants Use instead of The numeric coefficient should be a decimal with three significant figures A4 Find the final potential energy Find the potential energy of the particle with charge at the center of the original square A41 Formula for the electric potential energy The formula for the electric potential energy between charges and separated by distance is, where Express your answer in terms of,, and appropriate constants Use instead of The numeric coefficient should be a decimal with three significant figures

12 A5 Find the kinetic energy Using conservation of energy, find the initial kinetic energy A51 Conservation of energy of the particle with charge Conservation of energy implies that the sum of the initial kinetic and potential energies is equal to the sum of the final kinetic and potential energies Since the final kinetic energy, when the particle is momentarily at rest, is zero, you can express the initial kinetic energy in terms of the initial and final potential energies Express your answer in terms of,, and appropriate constants Use instead of The numeric coefficient should be a decimal with three significant figures A6 Formula for kinetic energy To obtain the inital velocity of the charged particle, you will need to recall the formula for the kinetic energy of a particle with mass and velocity : Express your answer in terms of,,, and appropriate constants Use instead of The numeric coefficient should be a decimal with three significant figures

13 Part B When the particle with charge reaches the center of the original square, it is, as stated in the problem, momentarily at rest Is the particle at equilibrium at that moment? B1 How to approach the problem Equilibrium means zero acceleration (not zero velocity) Zero acceleration, in turn, means that the net force applied to the particle must be zero Sketch the particle at its final position and the vectors representing the forces applied to it by the other three particles to estimate the net force on the particle yes no The exact value of the net force can be found by a calculation Potential of a Charged Annulus Description: Find the potential along the axis of a charged annulus An annular ring with a uniform surface charge density sits in the xy plane, with its center at the origin of the coordinate axes The annulus has an inner radius and outer radius Part A If you can find symmetries in a physical situation, you can often greatly simplify your calculations In this part you will find a symmetry in the annular ring before calculating the potential along the axis through the ring's center in Part B

14 through the ring's center in Part B Consider three sets of points: points lying on the vertical line A; those on circle B; and those on the horizontal line C, as shown in the figure Which set of points makes the same contribution toward the potential calculated at any point along the axis of the annulus? A1 Definition of the potential due to a point charge The potential due to a point charge at a distance from it is given by, where points on line A points on circle B points on line C Part B By exploiting the above symmetry, or otherwise, calculate the electric potential at a point on the axis of the annulus a distance from its center How to exploit the angular symmetry of the problem

15 B1 The total potential at a point on the axis of the annulus can be written as where is the distance from a point on the annulus to the point at which the potential is to be determined However, on account of the angular symmetry of this problem, it is more convenient to write this integral in terms of polar coordinates:, The integral over is easy and should be done first, since the integrand has no dependence on This will put the integral in the form where is the area of a thin annular slice of thickness and radius, B2 Find the area of an annular slice What is, the area of a thin annular slice of thickness and radius? Express your answer in terms of and Now do the integral over Don't forget that is a function of You will need to use a variable substitution B3 Doing the integral Set What is? Express your answer in terms of and? Substitute for and in terms of and to do the integral Find the new limits of integration

16 Substitute for and in terms of and to do the integral Find the new limits of integration carefully B4 A formula for the integral The integral has antiderivative Express your answer in terms of some or all of the variables,,, and Use It is interestering to note that the potential at any point on the axis of a disk of radius can be obtained from the expression above by setting and Doing so, one obtains Conversely, the annulus can be thought of as the superposition of two disks, one with charge density and radius, and the other with charge density and radius In the region from the center to, the opposite charge densities cancel out, so the net charge distribution would be just like that of the annulus Moreover, by adding the potentials due to these two disks, using the formula above, you would recover the potential of the annulus It is also instructive to look at the general behavior of these potentials as a function of the parameters Clearly, the potential increases with increasing charge densities, as well as with increasing areas (if

17 the charge density is held constant), which intuitively seems reasonable However, if the distance increases, it is not clear whether the potential should grow, since appears in both terms, of which one is subtracted from the other If you are far from the disk, the disk looks like a point, and the potential should drop off, just like the potential due to a point charge Indeed, on account of the negative second term in the expressions, this is the case Try some values or check that the derivative of is indeed negative You can also check that the above expression actually reduces to the potential due to a point charge for Charged Mercury Droplets Description: A large mercury drop, at known electric potential, breaks into n smaller drops Find the ratio of the electric potentials of the original and final drops A uniformly charged spherical droplet of mercury has electric potential The droplet then breaks into throughout the droplet identical spherical droplets, each of which has electric potential throughout its volume The not interact significantly Part A small droplets are far enough apart form one another that they do Find, the ratio of, the electric potential throughout the initial drop, to, the electric potential throughout one of the smaller drops A1 How to approach the problem Mercury is a metal and therefore a conductor It also has very high surface tension, so that droplets of mercury are roughly spherical The electric potential throughout the volume of a spherical conducting droplet with charge and radius is given by Write expressions for both electric potentials and Express all the quantities involved in terms of the charge and radius of the big drop and, and then compute the ratio A2 Find the charge on the small droplets Keeping conservation of charge in mind, find the charge on each small droplet Express your answer in terms of, the charge on the big droplet, and

18 A3 Find the radius of a small droplet What is the radius A31 Consider volume of each small droplet? The volume of each small spherical droplet is that of the large droplet In general, the volume of a sphere of radius is equal to Express your answer in terms of, the radius of the big droplet, and The ratio should be dimensionless and should depend only on ± The Geiger Counter Description: ± Includes Math Remediation Calculate the potential difference between the inner wire and the outer cylinder in a Geiger counter, given the electric field at a particular point A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path A thin wire lies on the axis of a hollow metal cylinder and is insulated from it

19 A large potential difference is established between the wire and the outer cylinder, with the wire at a higher potential; this sets up a strong electric field directed radially outward When ionizing radiation enters the device, it ionizes a few air molecules The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted into an audible click Suppose the radius of the central wire is 145 micrometers and the radius of the hollow cylinder is 180 centimeters Part A What potential difference between the wire and the cylinder produces an electric field of volts per meter at a distance of 120 centimeters from the axis of the wire? (Assume that the wire and cylinder are both very long in comparison to their radii) A1 How to approach the problem According to the introduction, and looking at the illustration, there is a positive charge on the inner wire and a negative charge of equal magnitude on the outer cylinder, which creates an electric field inside the cylinder that points radially outward, or away from the wire Therefore, the first thing to do is to write a generic equation for the electric field between the cylinder and the wire Using this, find a corresponding equation for the potential difference between the wire and cylinder Finally, combine the two equations and calculate the potential difference at the given point A2 Find an expression for the electric field Assume that the inner wire of a Geiger counter has a charge per unit length of, and the outer

20 cylinder has an equal but opposite charge per unit length, Find a generic expression for the electric field inside the Geiger counter as a function of the radial distance from the inner wire A21 Using Gauss's law The easiest way to solve this is to use Gauss's law Because of symmetrical properties of the wire, the Gaussian surface needed will be a cylinder with its ends perpendicular to the wire, and with arbitrary length and arbitrary radius, such that, where is the radius of the inner wire and that of the metal cylinder Again, because of symmetry, the electric field will have only a radial component, and only the flux through the side walls will contribute to the total flux through the cylinder Also note that since only the enclosed charge will give rise to the electric field, only the charge per unit length used in the calculations From Gauss's law, then, on the central wire will be, where the surface integral is calculated over the side walls of the Gaussian cylinder Use as the permittivity of free space and express your answer in terms of some or all the variables,, and any appropriate constants A3 Find an expression for the potential difference Consider a Geiger counter whose central wire has radius and outer cylinder has radius, and let be the charge per unit length on the central wire Using the equation for the electric field inside a Geiger counter found in 2, write a general expression for the potential difference between the inner wire and the outer cylinder Recall that Use, as the permittivity of free space and express your answer in terms of some or all the variables,,, and any appropriate constants

21 A4 Find a new expression for the electric field Using the equation for the potential difference that was found in 3, rewrite the expression for the in terms of electric field and the radial distance Express your answer in terms of some or all the variables,,,, and any appropriate constants By combining the equations for the electric field and the potential difference, the charge per unit length on the wire has been eliminated from the equations This is necessary, since no information about it is given in this problem A5 Putting it all together Information about the electric field at a particular point was given in the introduction, so the potential difference between the wire and the cylinder can now be calculated at that point Express your answer numerically in volts It is also possible to find the potential difference between the inner wire and the outer cylinder by integrating the electric field from to, and then from to, using the given value of the field at as an intermediate integration limit However, this is much more difficult to do, since the charge per unit length on the wire is not known, and is not necessary if you instead find an expression for the electric field in terms of the potential difference Speed of an Electron in an Electric Field

22 Description: Calculate the final speed of an electron released from rest between two stationary positive point charges of given magnitudes The distance between the stationary charges and the final distance between the electron and one of the positive charges are also given Two stationary positive point charges, charge 1 of magnitude 300 and charge 2 of magnitude 175, are separated by a distance of 540 An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges Part A What is the speed of the electron when it is 100 from charge 1? A1 How to approach the problem Only the final speed of the electron is needed, so an easy way to solve this problem is by using energy techniques Since the only force that acts on the electron is the conservative electric force, mechanical energy is conserved Thus, find the potential energy at the initial and final positions of the electron from the expression of the potential due to a collection of point charges Also, you can easily determine the initial kinetic energy of the electron, since it is released from rest Now you have enough information to write the equation of conservation of energy and calculate the speed of the electron from its final kinetic energy A2 Calculate the potential at the midpoint What is the potential A21 Electric potential at the midpoint between the two stationary positive charges? The potential due to a single point charge is, where , is the distance from the point charge to the point at which the potential is calculated, and is the charge If instead of a single point charge, there is a collection of point charges, the total potential is given by the sum of the potentials due to each charge A22 Find the potential due to charge 1 What is the potential at the midpoint due to charge 1 alone? Express your answer numerically in volts

23 For a collection of point charges, the total potential at any point is the sum of the potentials due to each charge Express your answer numerically in volts A3 Calculate the initial potential energy Calculate the potential energy of the electron at the midpoint between the two stationary positive charges, using the potential at that point A31 How to find the potential energy from the potential Recall that the potential energy associated with a test charge is related to the potential by the equation In this case, the test charge is the electron Express your answer numerically in joules A4 Calculate the initial kinetic energy Calculate the initial kinetic energy positive charges A41 Initial velocity of the electron of the electron at the midpoint between the two stationary Recall that the electron is initially at rest; thus its inital velocity is zero Express your answer numerically in joules

24 Express your answer numerically in joules A5 Calculate the final potential energy Calculate the potential energy of the electron at its final position, a point along the line connecting the positive stationary charges at distance 100 from charge 1 A51 Calculate the potential at the final position of the electron Calculate the potential at the final position of the electron, that is, at a point along the line connecting the positive stationary charges 100 from charge 1 Express your answer numerically in volts Express your answer in joules A6 Putting it all together Once you know the initial kinetic energy, the initial potential energy, and the final potential energy of the electron, you can calculate its final kinetic energy by applying conservation of energy: Finally, use the equation for the kinetic energy of a particle (in this case, the moving electron) to calculate its speed, since the mass of the electron is a known constant, Express your answer in meters per second

25 ANSWER : Note that the electric field between the two charges is not constant, so the easiest way to do these calculations is to use conservation of energy It is possible to integrate along the path of the electron, using the electric field as a function of the distance from each charge, but this is much more difficult to do and not necessary for the problem PROBLEMS: 294 Model: The mechanical energy of the charged particles is conserved A parallel-plate capacitor has a uniform electric field Visualize: The figure shows the before-and-after pictorial representation Solve: The potential energy is defined as U U 0 + qex, where x is the distance from the negative plate and U 0 is the potential energy at the negative plate (at x 0 m) Thus, the change in the potential energy of the proton as it moves from the positive plate to the negative plate is ΔU p U f U i U J ( ) ( U 0 + eed ) eed This decrease in potential energy appears as an increase in the proton s kinetic energy: ΔK K f K i 1 m v 2 2 p f p 1 m v 2 2 p i p 1 m v 2 2 p f p Applying the law of conservation of mechanical energy ΔK + ΔU p 0 J, we have 1 mv eed 2 f p ( ) 0 J v f p 2eEd m p When the proton is replaced with a helium ion and the same experiment is repeated, Dividing the two equations, 2 v f ion 2eEd m ion v f ion m p v m f p 1 4 ( 50,000 m/s) 25,000 m/s ion Assess: Being a heavier particle, the helium ion s velocity is expected to be smaller compared to the proton s velocity

26 2915 Model: Energy is conserved The potential energy is determined by the electric potential Visualize: The figure shows a before-and-after pictorial representation of a proton moving through a potential difference Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential (b) Using the conservation of energy equation, ΔV mv 2 i 2e K f + U f K i + U i K f + qv f K i + qv i V f V i 1 ( q K K i f ) ( 1 e) ( 1 mv 2 0 J 2 i ) ( )( 800,000 m/s) 2 2( C) kg 3340 V Assess: A positive ΔV confirms that the proton moves into a higher potential region 2930 Model: The net potential is the sum of the scalar potentials due to each charge Visualize: Solve: Let the point on the y-axis where the electric potential is zero be at a distance y from the origin At this point, V 1 + V 2 0 V This means 3 1 q 1 + q 2 r 2 r C 0 V + ( 90 cm) 2 + y C 0 ( 160 cm) 2 + y 2 ( 16 cm) 2 + y 2 4 ( 9 cm) 2 + y 2 9( 256 cm 2 + y ) 2 16( 81 cm 2 + y ) 2 7y cm 2 y ±12 cm

27 2945 Model: Energy is conserved The proton s potential energy inside the capacitor can be found from the capacitor s potential difference Visualize: Please refer to Figure P2945 Solve: (a) The electric potential at the midpoint of the capacitor is 250 V This is because the potential inside a parallel-plate capacitor is V Es where s is the distance from the negative electron The proton has charge q e and its potential energy at a point where the capacitor s potential is V is U ev The proton will gain potential energy ΔU eδv e(250 V) C (250 V) J if it moves all the way to the positive plate This increase in potential energy comes at the expense of kinetic energy which is ( )( 200,000 m/s) J K 1 2 mv kg This available kinetic energy is not enough to provide for the increase in potential energy if the proton is to reach the positive plate Thus the proton does not reach the plate because K < ΔU (b) The energy-conservation equation K f + U f K i + U i is ( ) 1 2 mv f2 + qv f 1 2 mv i2 + qv i 1 2 mv f2 1 2 mv i2 + q V i V f v f v 2 i + 2q ( m V V i f ) ( m/s) ( C) ( 250 V 0 V) kg m/s 2948 Model: Mechanical energy is conserved Visualize: Solve: The initial energy of the electron is the same as the energy at the turning point, when its speed is zero E i E f 1 2 m v 2 + U e i 1i + U 2i U 1f + U 2f ( )( e) 1 2 m v 1 e i2 + 2 e m 2 1 ( e) ( e) 1 ( kg) ( m/s) 2 2( Nm 2 /C ) ( C) m 2( Nm 2 /C ) ( C) 2 r J J Nm 2 r m nm r r

28 Since r nm ( ) 2 + y 2, y ( nm) 2 ( 0055 nm) nm Assess: The electron moves a distance outward that is less than the distance to each of the protons 2956 Model: Energy is conserved Visualize: The alpha particle is initially at rest (v i alpha 0 m/s) at the surface of the thorium nucleus The potential energy of the alpha particle is U i alpha After the decay, the alpha particle is far away from the thorium nucleus, U f alpha 0 J, and moving with speed v f alpha Solve: Initially, the alpha particle has potential energy and no kinetic energy As the alpha particle is detected in the laboratory, the alpha particle has kinetic energy but no potential energy Energy is conserved, so K f alpha + U f alpha K i alpha + U i alpha This equation is v f alpha 1 2 ( 360e ) mr i 1 mv J 0 J f alpha ( 2e) ( 90e) r i ( N m 2 /C 2 ) C ( ) 2 ( )( m) kg m/s 2964 Solve: A charged particle placed inside a uniformly charged spherical shell experiences no electric force That is, E 0 V/m inside the shell We know from Section 295 that a difference in potential between two points or two plates is the source of an electric field Since the potential on the surface of the shell is V Q R, the potential inside must be the same This ensures that the potential difference is zero and hence the electric field is zero inside the shell The potential at the center of the spherical shell is thus the same as at the surface That is, V center V surface Q R 2968 Model: The net potential is the sum of potentials from all the charges Visualize: Please refer to Figure P2968 Point P at which we want the net potential due to the linear electric quadrupole is far away compared to the separation s, that is, y >> s Solve: The net potential at P is V net V 1 + V 2 + V 3 1 q 1 y s + 1 q 2 y + 1 q 3 y + s At distances y >> s, ( ) ( ) ( ) 1 +q y s + 2q + +q y y + s q y 2 + ys 2y 2 + 2s 2 + y 2 ys y( y 2 s ) 2 q 2s 2 y y 2 s 2 ( )

29 V net 1 2 ( 2qs ) 1 y 3 where Q 2qs 2 is called the electric quadrupole moment of the charge distribution Assess: This charge distribution is in fact a combination of two dipoles As seen above their effects do not completely cancel Q y Model: The disk has a uniform surface charge density η Q A Q π ( R out R in ) Visualize: Please refer to Figure 2931 Orient the disk in the xy-plane, with point P at distance z Divide the disk into rings of equal width Δr Ring i has radius r i and charge ΔQ i Solve: Using the result of Example 2911, we write the potential at distance z of ring i as Noting that ΔQ i ηδa i η2πr i Δr, V i 1 ΔQ i V V r 2 i + z 2 i 1 i i ΔQ i r i 2 + z 2 V 1 η2π 2πε 0 Q In the limit R in 0 m, R 2 2 ( out R in ) i r i Δr r i 2 + z 2 2 R out η 2ε 0 R out R in r dr r 2 + z 2 + z 2 R 2 in + z 2 Q V 2πε 0 R out η r 2 + z 2 2ε 0 2 R 2 out R out Rin + z 2 z This is the same result obtained for a disk of charge in Example 2912 η 2ε 0 2 R out + z 2 R 2 in + z Model: Assume the wire is a line of charge with uniform linear charge density The electric potential at the point is the sum of contributions from all charges present Visualize: Please refer to Figure CP2983 Let the origin be the center of our coordinate system Divide the charged wire into two straight sections and the central semicircle Divide each section into N small segments, each of length Δx and with charge Δq Segment i, located at position x i, contributes a small amount of potential V i at the semicircle center Solve: For the right hand straight section, the contribution of the ith segment is V i Δq Δq λδx r i x i x i where Δq λδx The V i are now summed and the sum is converted to an integral giving 3R ( ) V straight section λ dx λ x 4πε R 0 R ln x R 3R λ R ln ( 3 ) The integration limits are set by the physical location of the straight section By symmetry, the left hand straight section adds the same amount to the total potential at the semicircle center The potential due to the ith segment of the semicircular section is V i ( ) Δq Δq r i R λ R Δθ R λ R Δθ where we have used the arc length Δx R Δθ The V i are now summed and the sum converted to an integral giving

30 λ π V semicircle dθ R λπ 0 R λ 4ε 0 R The total potential is the sum of the potentials due to the three sections: V V semicircle + 2V straight section λ 4ε 0 R + 2 λ R ln ( 3 ) λ 2ln(3) 1+ 4ε 0 R π