# Chapter 3. Special Techniques for Calculating Potentials. r ( r ' )dt ' ( ) 2

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1 Chater 3. Secial Techniues for Calculating Potentials Given a stationary charge distribution r( r ) we can, in rincile, calculate the electric field: E ( r ) Ú Dˆ r Dr r ( r ' )dt ' 2 where Dr r '-r. This integral involves a vector as an integrand and is, in general, difficult to calculate. In most cases it is easier to evaluate first the electrostatic otential V which is defined as V( r ) Ú r r ' Dr dt ' since the integrand of the integral is a scalar. The corresonding electric field E can then be obtained from the gradient of V since E - V The electrostatic otential V can only be evaluated analytically for the simlest charge configurations. In addition, in many electrostatic roblems, conductors are involved and the charge distribution r is not known in advance (only the total charge on each conductor is known). A better aroach to determine the electrostatic otential is to start with Poisson's euation 2 V - r e Very often we only want to determine the otential in a region where r. In this region Poisson's euation reduces to Lalace's euation 2 V There are an infinite number of functions that satisfy Lalace's euation and the aroriate solution is selected by secifying the aroriate boundary conditions. This Chater will concentrate on the various techniues that can be used to calculate the solutions of Lalace's euation and on the boundary conditions reuired to uniuely determine a solution. - -

2 3.. Solutions of Lalace's Euation in One-, Two, and Three Dimensions 3... Lalace's Euation in One Dimension In one dimension the electrostatic otential V deends on only one variable x. The electrostatic otential V(x) is a solution of the one-dimensional Lalace euation The general solution of this euation is d 2 V dx 2 Vx sx + b where s and b are arbitrary constants. These constants are fixed when the value of the otential is secified at two different ositions. Examle Consider a one-dimensional world with two oint conductors located at x m and at x m. The conductor at x m is grounded (V V) and the conductor at x m is ket at a constant otential of 2 V. Determine V. and The boundary conditions for V are V( ) b V V( ) s + b 2V The first boundary condition shows that b V. The second boundary condition shows that s 2 V/m. The electrostatic otential for this system of conductors is thus Vx 2x The corresonding electric field can be obtained from the gradient of V - dv( x) Ex dx -2 V / m - 2 -

3 The boundary conditions used here, can be used to secify the electrostatic otential between x m and x m but not in the region x < m and x > m. If the solution obtained here was the general solution for all x, then V would aroach infinity when x aroaches infinity and V would aroach minus infinity when x aroaches minus infinity. The boundary conditions therefore rovide the information necessary to uniuely define a solution to Lalace's euation, but they also define the boundary of the region where this solution is valid (in this examle m < x < m). The following roerties are true for any solution of the one-dimensional Lalace euation: Proerty : V(x) is the average of V(x + R) and V(x - R) for any R as long as x + R and x - R are located in the region between the boundary oints. This roerty is easy to roof: + Vx ( - R) Vx + r 2 + b + sx ( - R) + b sx + R 2 sx + b Vx This roerty immediately suggests a owerful analytical method to determine the solution of Lalace's euation. If the boundary values of V are and Vx ( a ) V a Vx ( b ) V b then roerty can be used to determine the value of the otential at (a + b)/2: V x a + b ˆ Ë 2 [ 2 V + V a b ] Next we can determine the value of the otential at x (3 a + b)/4 and at x (a + 3 b)/4 : V x 3a + b ˆ Ë 2 È 2 Î Vx a + V x a + b Ë 2 ˆ È V + a 2 V b Î V x a + 3b ˆ Ë 2 È 2 V x a + b ˆ Î Ë 2 + Vx b È 2 2 V + 3 a 2 V b Î - 3 -

4 This rocess can be reeated and V can be calculated in this manner at any oint between x a and x b (but not in the region x > b and x < a). Proerty 2: The solution of Lalace's euation can not have local maxima or minima. Extreme values must occur at the end oints (the boundaries). This is a direct conseuence of roerty. Proerty 2 has an imortant conseuence: a charged article can not be held in stable euilibrium by electrostatic forces alone (Earnshaw's Theorem). A article is in a stable euilibrium if it is located at a osition where the otential has a minimum value. A small dislacement away from the euilibrium osition will increase the electrostatic otential of the article, and a restoring force will try to move the article back to its euilibrium osition. However, since there can be no local maxima or minima in the electrostatic otential, the article can not be held in stable euilibrium by just electrostatic forces Lalace's Euation in Two Dimensions In two dimensions the electrostatic otential deends on two variables x and y. Lalace's euation now becomes 2 V x V y 2 This euation does not have a simle analytical solution as the one-dimensional Lalace euation does. However, the roerties of solutions of the one-dimensional Lalace euation are also valid for solutions of the two-dimensional Lalace euation: Proerty : The value of V at a oint (x, y) is eual to the average value of V around this oint Vx,y 2R Ú Circle VRdf where the ath integral is along a circle of arbitrary radius, centered at (x, y) and with radius R. Proerty 2: V has no local maxima or minima; all extremes occur at the boundaries

5 3..3. Lalace's Euation in Three Dimensions In three dimensions the electrostatic otential deends on three variables x, y, and z. Lalace's euation now becomes 2 V x V y V z 2 This euation does not have a simle analytical solution as the one-dimensional Lalace euation does. However, the roerties of solutions of the one-dimensional Lalace euation are also valid for solutions of the three-dimensional Lalace euation: Proerty : The value of V at a oint (x, y, z) is eual to the average value of V around this oint Vx,y,z 4R 2 Ú Shere VR 2 sin d df where the surface integral is across the surface of a shere of arbitrary radius, centered at (x,y,z) and with radius R. y P R d z x r Figure 3.. Proof of roerty

6 To roof this roerty of V consider the electrostatic otential generated by a oint charge located on the z axis, a distance r away from the center of a shere of radius R (see Figure 3.). The otential at P, generated by charge, is eual to V P d where d is the distance between P and. Using the cosine rule we can exress d in terms of r, R and d 2 r 2 + R 2-2rR cos The otential at P due to charge is therefore eual to V P r 2 + R 2-2rR cos The average otential on the surface of the shere can be obtained by integrating V P across the surface of the shere. The average otential is eual to V average 4R 2 Ú V P R 2 sin d df 4 Shere Ú 2 sin d r 2 + R 2-2rR cos 8e r 2 + R 2-2rR cos rr 8e Ë r + R rr - r - R rr ˆ r which is eual to the otential due to at the center of the shere. Alying the rincile of suerosition it is easy to show that the average otential generated by a collection of oint charges is eual to the net otential they roduce at the center of the shere. Proerty 2: The electrostatic otential V has no local maxima or minima; all extremes occur at the boundaries. Examle: Problem 3.3 Find the general solution to Lalace's euation in sherical coordinates, for the case where V deends only on r. Then do the same for cylindrical coordinates. Lalace's euation in sherical coordinates is given by - 6 -

7 r 2 r r V 2 ˆ Ë r + r 2 sin V ˆ sin Ë + 2 V r 2 sin 2 f 2 If V is only a function of r then and V V f Therefore, Lalace's euation can be rewritten as r 2 r r V 2 ˆ Ë r The solution V of this second-order differential euation must satisfy the following first-order differential euation: This differential euation can be rewritten as r 2 V r a constant V r a r 2 The general solution of this first-order differential euation is Vr - a r + b where b is a constant. If V at infinity then b must be eual to zero, and conseuently Vr - a r Lalace's euation in cylindrical coordinates is - 7 -

8 r r r V ˆ Ë r + 2 V r 2 f V z 2 If V is only a function of r then V f and Therefore, Lalace's euation can be rewritten as V z r r r V ˆ Ë r The solution V of this second-order differential euation must satisfy the following first-order differential euation: This differential euation can be rewritten as r V r a constant V r a r The general solution of this first-order differential euation is Vr a ln( r ) + b where b is a constant. The constants a and b are determined by the boundary conditions Uniueness Theorems Consider a volume (see Figure 3.2) within which the charge density is eual to zero. Suose that the value of the electrostatic otential is secified at every oint on the surface of this volume. The first uniueness theorem states that in this case the solution of Lalace's euation is uniuely defined

9 Boundary Volume Figure 3.2. First Uniueness Theorem To roof the first uniueness theorem we will consider what haens when there are two solutions V and V 2 of Lalace's euation in the volume shown in Figure 3.2. Since V and V 2 are solutions of Lalace's euation we know that and 2 V 2 V 2 Since both V and V 2 are solutions, they must have the same value on the boundary. Thus V V 2 on the boundary of the volume. Now consider a third function V 3, which is the difference between V and V 2 V 3 V -V 2 The function V 3 is also a solution of Lalace's euation. This can be demonstrated easily: 2 V 3 2 V - 2 V 2 The value of the function V 3 is eual to zero on the boundary of the volume since V V 2 there. However, roerty 2 of any solution of Lalace's euation states that it can have no local maxima or minima and that the extreme values of the solution must occur at the boundaries. Since V 3 is a solution of Lalace's euation and its value is zero everywhere on the boundary of the volume, the maximum and minimum value of V 3 must be eual to zero. Therefore, V 3 must be eual to zero everywhere. This immediately imlies that - 9 -

10 V V 2 everywhere. This roves that there can be no two different functions V and V 2 that are solutions of Lalace's euation and satisfy the same boundary conditions. Therefore, the solution of Lalace's euation is uniuely determined if its value is a secified function on all boundaries of the region. This also indicates that it does not matter how you come by your solution: if (a) it is a solution of Lalace's euation, and (b) it has the correct value on the boundaries, then it is the right and only solution. Boundary Figure 3.3. System with conductors. The first uniueness theorem can only be alied in those regions that are free of charge and surrounded by a boundary with a known otential (not necessarily constant). In the laboratory the boundaries are usually conductors connected to batteries to kee them at a fixed otential. In many other electrostatic roblems we do not know the otential at the boundaries of the system. Instead we might know the total charge on the various conductors that make u the system (note: knowing the total charge on a conductor does not imly a knowledge of the charge distribution r since it is influenced by the resence of the other conductors). In addition to the conductors that make u the system, there might be a charge distribution r filling the regions between the conductors (see Figure 3.3). For this tye of system the first uniueness theorem does not aly. The second uniueness theorem states that the electric field is uniuely determined if the total charge on each conductor is given and the charge distribution in the regions between the conductors is known. - -

11 The roof of the second uniueness theorem is similar to the roof of the first uniueness theorem. Suose that there are two fields E and E 2 that are solutions of Poisson's euation in the region between the conductors. Thus E r e and E 2 r e where r is the charge density at the oint where the electric field is evaluated. The surface integrals of E and E 2, evaluated using a surface that is just outside one of the conductors with charge Q i, are eual to Q i / e. Thus Ú Surface conductor i E da Q i e Ú Surface conductor i E 2 da Q i e The difference between E and E 2, E 3 E - E 2, satisfies the following euations: E 3 E - E 2 r e - r e Ú E 3 da Ú E da - Ú E 2 da Surface conductor i Surface conductor i Surface conductor i Q i e - Q i e Consider the surface integral of E 3, integrated over all surfaces (the surface of all conductors and the outer surface). Since the otential on the surface of any conductor is constant, the electrostatic otential associated with E and E 2 must also be constant on the surface of each conductor. Therefore, V 3 V -V 2 will also be constant on the surface of each conductor. The surface integral of V 3 E 3 over the surface of conductor i can be written as Ú V 3 E 3 da V 3 Ú E 3 da Surface conductor i Surface conductor i - -

12 Since the surface integral of V 3 E 3 over the surface of conductor i is eual to zero, the surface integral of V 3 E 3 over all conductor surfaces will also be eual to zero. The surface integral of V 3 E 3 over the outer surface will also be eual to zero since V 3 on this surface. Thus Ú V 3 E 3 da All surfaces The surface integral of V 3 E 3 can be rewritten using Green's identity as Ú V 3 E 3 da - Ú V 3 V 3 da - Ú ( V 3 2 V 3 + ( V 3 ) ( V 3 ))dt All surfaces All surfaces Volume between conductors - Ú (-V 3 ( E 3 ) + E 3 E 3 )dt - Ú E 2 3 dt Volume between conductors Volume between conductors where the volume integration is over all sace between the conductors and the outer surface. Since E 2 3 is always ositive, the volume integral of E 2 3 can only be eual to zero if E 2 3 everywhere. This imlies immediately that E E 2 everywhere, and roves the second uniueness theorem Method of Images Consider a oint charge held as a distance d above an infinite grounded conducting lane (see Figure 3.4). The electrostatic otential of this system must satisfy the following two boundary conditions: Vx,y, Vx,y,z Æ when Ï x Æ Ô Ì y Æ Ô Ó z Æ A direct calculation of the electrostatic otential can not be carried out since the charge distribution on the grounded conductor is unknown. Note: the charge distribution on the surface of a grounded conductor does not need to be zero

13 z axis d Figure 3.4. Method of images. Consider a second system, consisting of two oint charges with charges + and -, located at z d and z -d, resectively (see Figure 3.5). The electrostatic otential generated by these two charges can be calculated directly at any oint in sace. At a oint P (x, y, ) on the xy lane the electrostatic otential is eual to Vx,y, È Î x 2 + y 2 + d x 2 + y 2 + d 2 z axis d P d - Figure 3.5. Charge and image charge. The otential of this system at infinity will aroach zero since the otential generated by each charge will decrease as /r with increasing distance r. Therefore, the electrostatic otential - 3 -

14 generated by the two charges shown in Figure 3.5 satisfies the same boundary conditions as the system shown in Figure 3.4. Since the charge distribution in the region z > (bounded by the xy lane boundary and the boundary at infinity) for the two systems is identical, the corollary of the first uniueness theorem states that the electrostatic otential in this region is uniuely defined. Therefore, if we find any function that satisfies the boundary conditions and Poisson's euation, it will be the right answer. Consider a oint (x, y, z) with z >. The electrostatic otential at this oint can be calculated easily for the charge distribution shown in Figure 3.5. It is eual to Vx,y,z È Î x 2 + y z - d x 2 + y 2 + z + d Since this solution satisfies the boundary conditions, it must be the correct solution in the region z > for the system shown in Figure 3.4. This techniue of using image charges to obtain the electrostatic otential in some region of sace is called the method of images. The electrostatic otential can be used to calculate the charge distribution on the grounded conductor. Since the electric field inside the conductor is eual to zero, the boundary condition for E (see Chater 2) shows that the electric field right outside the conductor is eual to E Outside s n ˆ s k ˆ e e where s is the surface charge density and n ˆ is the unit vector normal to the surface of the conductor. Exressing the electric field in terms of the electrostatic otential V we can rewrite this euation as V s e E z -e z z Substituting the solution for V in this euation we find s - 4 È -( z - d ) ( x 2 + y 2 + ( z - d ) 2 ) 3/2 + Î ( z + d) ( x 2 + y 2 + ( z + d ) 2 ) 3/2 - z 2 d x 2 + y 2 + d 2 3/2 Only in the last ste of this calculation have we substituted z. The induced charge distribution is negative and the charge density is greatest at (x, y, z ). The total charge on the conductor can be calculated by surface integrating of s: - 4 -

15 Q total 2 Ú s da Ú Ú s ( r )rdrd Surface where r 2 x 2 + y 2. Substituting the exression for s in the integral we obtain Q total -dú r 2 + d 2 3/2 rdr d r 2 + d 2 È d - Î d - As a result of the induced surface charge on the conductor, the oint charge will be attracted towards the conductor. Since the electrostatic otential generated by the charge imagecharge system is the same as the charge-conductor system in the region where z >, the associated electric field (and conseuently the force on oint charge ) will also be the same. The force exerted on oint charge can be obtained immediately by calculating the force exerted on the oint charge by the image charge. This force is eual to ˆ 2 k F - 2 2d There is however one imortant difference between the image-charge system and the real system. This difference is the total electrostatic energy of the system. The electric field in the image-charge system is resent everywhere, and the magnitude of the electric field at (x, y, z) will be the same as the magnitude of the electric field at (x, y, -z). On the other hand, in the real system the electric field will only be non zero in the region with z >. Since the electrostatic energy of a system is roortional to the volume integral of E 2 the electrostatic energy of the real system will be /2 of the electrostatic energy of the image-charge system (only /2 of the total volume has a non-zero electric field in the real system). The electrostatic energy of the image-charge system is eual to W image 2 2d The electrostatic energy of the real system is therefore eual to W real 2 W image - 2 4d The electrostatic energy of the real system can also be obtained by calculating the work reuired to be done to assemble the system. In order to move the charge to its final osition we will - 5 -

16 have to exert a force oosite to the force exerted on it by the grounded conductor. The work done to move the charge from infinity along the z axis to z d is eual to W real d 2 Ú 4z 2 dz - 2 4z d - 2 4d which is identical to the result obtained using the electrostatic otential energy of the imagecharge system. Examle: Examle Problem 3.7 A oint charge is situated a distance s from the center of a grounded conducting shere of radius R (see Figure 3.6). a) Find the otential everywhere. b) Find the induced surface charge on the shere, as function of. Integrate this to get the total induced charge. c) Calculate the electrostatic energy of this configuration. R V s Figure 3.6. Examle Problem 3.7. a) Consider a system consisting of two charges and ', located on the z axis at z s and z z', resectively. If the otential roduced by this system is identical everywhere to the otential roduced by the system shown in Figure 3.6 then the osition of oint charge ' must be chosen such that the otential on the surface of a shere of radius R, centered at the origin, is eual to zero (in this case the boundary conditions for the otential generated by both systems are identical). We will start with determining the correct osition of oint charge '. The electrostatic otential at P (see Figure 3.7) is eual to - 6 -

17 This euation can be rewritten as V P s - R + ' - s - R R - z' ' R - z' V Q z' R d' P' ' P d s The electrostatic otential at Q is eual to This euation can be rewritten as Figure 3.7. Image-charge system. V Q Combining the two exression for ' we obtain or s + R + ' - s + R R + z' ' R + z' s - R ( R - z' ) s + R ( R + z' ) ( s + R) ( R - z' ) ( s - R) ( R + z' ) This euation can be rewritten as z' ( s - R) + z' ( s + R) 2sz ' Rs ( + R) - Rs ( - R) 2R 2-7 -

18 The osition of the image charge is eual to z' R2 s The value of the image charge is eual to ' - s + R R + z' - Á s + R Ë R + ˆ - R s s R2 Now consider an arbitrary oint P' on the circle. The distance between P' and charge is d and the distance between P' and charge ' is eual to d'. Using the cosine rule (see Figure 3.7) we can exress d and d' in terms of R, s, and : d R 2 + s 2-2Rscos d' R 2 + z' 2-2Rz'cos R 2 + R4 s 2-2R R2 s cos The electrostatic otential at P' is eual to V P' È Î d + È ' d' Î - R 2 + s 2-2Rscos + R 2 + R4 s 2 R s R2-2R s cos È Î R 2 + s 2-2Rscos - R 2 + s 2-2Rscos Thus we conclude that the configuration of charge and image charge roduces an electrostatic otential that is zero at any oint on a shere with radius R and centered at the origin. Therefore, this charge configuration roduces an electrostatic otential that satisfies exactly the same boundary conditions as the otential roduced by the charge-shere system. In the region outside the shere, the electrostatic otential is therefore eual to the electrostatic otential roduced by the charge and image charge. Consider an arbitrary oint ( r,,f ). The distance between this oint and charge is d and the distance between this oint and charge ' is eual to d'. These distances can be exressed in terms of r, s, and using the cosine rule: d r 2 + s 2-2rs cos - 8 -

19 d' r 2 + z' 2-2rz 'cos r 2 + R4 s 2-2r R2 s cos The electrostatic otential at ( r,,f ) will therefore be eual to Vr, (,f ) È Î d + ' d' { - r 2 + s 2-2rs cos + r 2 + R4 s 2 R s R2-2r s cos } { r 2 + s 2-2rs cos - } 2 rsˆ + R 2-2rs cos Ë R b) The surface charge density s on the shere can be obtained from the boundary conditions of E E outside - E inside E outside s n ˆ s r ˆ e e where we have used the fact that the electric field inside the shere is zero. This euation can be rewritten as V s e E r -e r Substituting the general exression for V into this euation we obtain s - 4 { -r + scos r 2 + s 2-2rs cos 3/2 - rsˆ Á Ë Ë R - rs2 + scos 2 R 2 ˆ + R 2-2rs cos 3/2 } r R - 4 { -R + scos R 2 + s 2-2Rscos - s2 R + scos s 2 + R 2-2Rscos - 3/2 3/2} - s 2 - R 2 4R ( R 2 + s 2-2Rscos ) 3/2-9 -

20 The total charge on the shere can be obtained by integrating s over the surface of the shere. The result is Q Ú sr 2 sin d df - 2 Rs2 - R 2 sin d Ú R 2 + s 2-2Rscos 3/2-2 Rs2 - R 2 È Î - sr R 2 + s 2-2Rscos s 2 - R 2 È 2 s Î R + s - R - s - R s c) To obtain the electrostatic energy of the system we can determine the work it takes to assemble the system by calculating the ath integral of the force that we need to exert in charge in order to move it from infinity to its final osition (z s). Charge will feel an attractive force exerted by the induced charge on the shere. The strength of this force is eual to the force on charge exerted by the image charge '. This force is eual to F ' ' s - z' 2 ˆ k - R Ë s ˆ Á Ë s - R2 s ˆ k - sr 2 s 2 - R 2 2 ˆ ˆ 2 k The force that we must exert on to move it from infinity to its current osition is oosite to F '. The total work reuired to move the charge is therefore eual to s W Ú -F ' dl s zr 2 Ú -R 2 z 2 - R 2 2 z 2 - R 2 2 dz s - 8e R 2 s 2 - R 2 Examle: Problem 3. Two semi-infinite grounded conducting lanes meet at right angles. In the region between them, there is a oint charge, situated as shown in Figure 3.8. Set u the image configuration, and calculate the otential in this region. What charges do you need, and where should they be located? What is the force on? How much work did it take to bring in from infinity? Consider the system of four charges shown in Figure 3.9. The electrostatic otential generated by this charge distribution is zero at every oint on the yz lane and at every oint on the xz lane. Therefore, the electrostatic otential generated by this image charge distribution satisfies the same boundary conditions as the electrostatic otential of the original system. The - 2 -

21 otential generated by the image charge distribution in the region where x > and y > will be identical to the otential of the original system. The otential at a oint P (x, y, z) is eual to V P ( x - a ) 2 + y - b z 2 - ( x - a ) 2 + y + b 2 + z x + a 2 + y - b 2 + z 2 + ( x + a ) 2 + ( y + b ) 2 + z 2 y a V b x V Figure 3.8. Problem a y a b b b b x a a - Figure 3.9. Image charges for roblem 3.. The force exerted on can be obtained by calculating the force exerted on by the image charges. The total force is eual to the vector sum of the forces exerted by each of the three - 2 -

22 image charges. The force exerted by the image charge located at (-a, b, ) is directed along the negative x axis and is eual to F - 2 4a ˆ 2 i The force exerted by the image charge located at (a, -b, ) is directed along the negative y axis and is eual to F 2-2 4b ˆ 2 j The force exerted by the image charge located at (-a, -b, ) is directed along the vector connecting (-a, -b, ) and (a, b, ) and is eual to F 3 2 4a 2 + 4b 2 aˆ i + bˆ j a 2 + b 2 2 6e a 2 + b 2 3/2 aˆ ( i + bˆ j ) The total force on charge is the vector sum of F, F 2 and F 3 : F tot F + F 2 + F 3-2 6e Á Á a 2 - Ë Ë a a 2 + b 2 3/2 ˆ ˆ i + Á b 2 - Ë b a 2 + b 2 3/2 ˆ ˆ ˆ j The electrostatic otential energy of the system can, in rincile, be obtained by calculating the ath integral of -F tot between infinity and (a, b, ). However, this is not trivial since the force -F tot is a rather comlex function of a and b. An easier techniue is to calculate the electrostatic otential energy of the system with charge and image charges. The otential energy of this system is eual to W image - 2 2a b + 2 ˆ Á 2 Á Ë 4a 2 + 4b 2 8e Ë a 2 + b 2 - a - ˆ b However, in the real system the electric field is only non-zero in the region where x > and y >. Therefore, the total electrostatic otential energy of the real system is only /4 of the total electrostatic otential energy of the image charge system. Thus W real 4 W 2 image Á 32e Ë a 2 + b 2 - a - ˆ b

23 3.3. Searation of Variables Searation of variables: Cartesian coordinates A owerful techniue very freuently used to solve artial differential euations is searation of variables. In this section we will demonstrate the ower of this techniue by discussing several examles. Examle: Examle 3.3 (Griffiths) Two infinite, grounded, metal lates lie arallel to the xz lane, one at y, the other at y (see Figure 3.). The left end, at x, is closed off with an infinite stri insulated from the two lates and maintained at a secified otential V ( y). Find the otential inside this "slot". y z V(y) V V x Figure 3.. Examle 3.3 (Griffiths). The electrostatic otential in the slot must satisfy the three-dimensional Lalace euation. However, since V does not have a z deendence, the three-dimensional Lalace euation reduces to the two-dimensional Lalace euation: 2 V x V y 2 The boundary conditions for the solution of Lalace's euation are:. V(x, y ) (grounded bottom late)

24 2. V(x, y ) (grounded to late). 3. V(x, y) V (y) (late at x ). 4. V Æ when x Æ. These four boundary conditions secify the value of the otential on all boundaries surrounding the slot and are therefore sufficient to uniuely determine the solution of Lalace's euation inside the slot. Therefore, if we find one solution of Lalace's euation satisfying these boundary conditions than it must be the correct one. Consider solutions of the following form: Vx,y Xx Y( y) If this is a solution of the two-dimensional Lalace euation than we must reuire that 2 [ x 2 Xx Y( y) ] + 2 y 2 This euation can be rewritten as [ Xx Y( y) ] Y y 2 Xx Xx x 2 + Y y 2 Xx x 2 2 Yy y 2 + Xx 2 Yy y 2 The first term of the left-hand side of this euation deends only on x while the second term deends only on y. Therefore, if this euation must hold for all x and y in the slot we must reuire that Xx 2 Xx x 2 C constant and Yy 2 Yy y 2 -C The differential euation for X can be rewritten as 2 Xx x 2 -C Xx

25 If C is a negative number than this euation can be rewritten as 2 Xx x 2 + k 2 Xx where k 2 -C. The most general solution of this euation is Xx A cos( kx) + B sin( kx) However, this function is an oscillatory function and does not satisfy boundary condition # 4, which reuires that V aroaches zero when x aroaches infinity. We therefore conclude that C can not be a negative number. If C is a ositive number then the differential euation for X can be written as The most general solution of this euation is 2 Xx x 2 - k 2 Xx Xx Ae kx + Be -kx This solution will aroach zero when x aroaches infinity if A. Thus Xx Be -kx The solution for Y can be obtained by solving the following differential euation: 2 Yy y 2 + k 2 Yy The most general solution of this euation is Yy C cos( ky) + D sin( ky) Therefore, the general solution for the electrostatic otential V(x,y) is eual to e -kx C cos( ky) + D sin( ky) Vx,y

26 where we have absorbed the constant B into the constants C and D. The constants C and D must be chosen such that the remaining three boundary conditions (, 2, and 3) are satisfied. The first boundary condition reuires that V(x, y ) : e -kx ( C cos( ) + D sin( ) ) Ce -kx Vx,y which reuires that C. The second boundary condition reuires that V(x, y ) : Vx,y ( ) De -kx sin( k ) which reuires that sin k integers:. This condition limits the ossible values of k to ositive k,2,3,4,... Note: negative values of k are not allowed since ex(-kx) aroaches zero at infinity only if k >. To satisfy boundary condition # 3 we must reuire that Vx (,y) D sin( ky) V ( y) This last exression suggests that the only time at which we can find a solution of Lalace's euation that satisfies all four boundary conditions has the form ex kx sin ky is when V ( y) haens to have the form sin ky. However, since k can take on an infinite number of values, there will be an infinite number of solutions of Lalace's euation satisfying boundary conditions #, # 2 and # 4. The most general form of the solution of Lalace's euation will be a linear suerosition of all ossible solutions. Thus D k e -kx sin( ky) Vx,y Boundary condition # 3 can now be written as Vx,y k D k sin( ky) k V ( y) Multilying both sides by sin(ny) and integrating each side between y and y we obtain sin( ny)sin( ky)dy sin( ny)v ( y)dy D k Ú k Ú

27 The integral on the left-hand side of this euation is eual to zero for all values of k excet k n. Thus Ú sin( ny)sin( ky)dy D k 2 d kn D k k k 2 D n The coefficients D k can thus be calculated easily: D k 2 Ú sin( ky)v ( y)dy The coefficients D k are called the Fourier coefficients of V ( y). The solution of Lalace's euation in the slot is therefore eual to where D k e -kx sin( ky) Vx,y k D k 2 Ú sin( ky)v ( y)dy Now consider the secial case in which V ( y) constant V. In this case the coefficients D k are eual to D k k 2 sin( ky)v Ú ( y)dy - 2 V cos ky 2 V k Ï Ô if k is even ( - cos ( k )) Ô Ì Ô 4 Ô Ó V k if k is odd The solution of Lalace's euation is thus eual to Vx,y 4V k e -kx sin ky k,3,5,.. Examle: Problem 3.2 Find the otential in the infinite slot of Examle 3.3 (Griffiths) if the boundary at x consists to two metal stries: one, from y to y /2, is held at constant otential V, and the other, from y /2 to y is at otential -V

28 The boundary condition at x is V,y Ï V for < y < /2 Ô Ì Ô Ó -V for /2< y < The Fourier coefficients of the function V ( y) are eual to D k 2 Ú sin( ky)v ( y)dy 2 V /2 Ú sin( ky)dy - 2 V Ú sin( ky)dy /2 2 V k + cos k Ë Á Ë - 2cos 2 k ˆ ˆ 2 V k C k The values for the first four C coefficients are C C 3 C 2 4 C 4 It is easy to see that C k + 4 C k and therefore we conclude that The Fourier coefficients C k are thus eual to The electrostatic otential is thus eual to Ï 4 for k 2,6,,... Ô C k Ì Ô Ó otherwise Ï 8V for k 2,6,,... Ô k D k Ì Ô otherwise Ó Ô Vx,y 8V k 2,6,,... k e -kx sin( ky) Examle: Problem 3.3 For the infinite slot (Examle 3.3 Griffiths) determine the charge density s y x, assuming it is a conductor at constant otential V. on the stri at

29 The electrostatic otential in the slot is eual to Vx,y 4V k e -kx sin ky k,3,5,.. The charge density at the late at x can be obtained using the boundary condition for the electric field at a boundary: E x + - E x - E x + s e ˆ n where n ˆ is directed along the ositive x axis. Since E - V this boundary condition can be rewritten as Differentiating V(x,y) with resect to x we obtain At the x boundary we obtain V x -4V V x x + - s e V - 4V x x + e -kx sin( ky) k,3,5,... sin( ky) k,3,5,... The charge density s on the x stri is therefore eual to V s -e 4V e x x + sin( ky) k,3,5,... Examle: Double infinite slots The slot of examle 3.3 in Griffiths and its mirror image at negative x are searated by an insulating stri at x. If the charge density s(y) on the dividing stri is given, determine the otential in the slot. The boundary condition at x reuires that

30 E x + - E x - 2E x + s e ˆ n where n ˆ is directed along the ositive x axis. Here we have used the symmetry of the configuration which reuires that the electric field in the region x < is the mirror image of the field in the region x >. Since E - V this boundary condition can be rewritten as V - s ( y) x x + 2 We will first determine the otential in the x > region. Following the same rocedure as in Examle 3 we obtain for the electrostatic otential Vx,y e D k e -kx sin( ky) k where the constants D k must be chosen such that the boundary condition at x is satisfied. This reuires that V - x x + k kd k e -kx sin( ky) x + -kd k sin ky k - 2 s y e Thus kd k sin( ky) s y 2 k e The constants D k can be determined by multilying both sides of this euation with sin( my ) and integrating both sides with resect to y between y and y. The result is The constants C k are thus eual to s ( y)sin( my )dy 2e Ú kd k Ú sin( ky)sin( my)dy k D k e k The electrostatic otential is thus eual to Ú s ( y)sin( ky)dy md m 2-3 -

31 Vx,y È s ( y)sin( ky)dy e k { Ú }e -kx sin( ky) Î k Searation of variables: sherical coordinates Consider a sherical symmetric system. If we want to solve Lalace's euation it is natural to use sherical coordinates. Assuming that the system has azimuthal symmetry ( V / f ) Lalace's euation reads r 2 r r V 2 ˆ Ë r Multilying both sides by r 2 we obtain + r 2 sin V ˆ sin Ë r r V 2 ˆ Ë r + sin V ˆ sin Ë Consider the ossibility that the general solution of this euation is the roduct of a function Rr, which deends only on the distance r, and a function a( ), which deends only on the angle : Vr, Rr a( ) Substituting this "solution" into Lalace's euation we obtain a( ) r r R r ˆ 2 Ë Á r + Rr sin Dividing each term of this euation by Rr a we obtain Rr r r R r ˆ 2 Ë Á r + a a ˆ sin Ë Á sin a ˆ sin Ë Á The first term in this exression deends only on the distance r while the second term deends only on the angle. This euation can only be true for all r and if and Rr r r R r ˆ Á 2 mm + Ë r constant - 3 -

32 a ˆ sin a( ) sin Ë Á -mm + Consider a solution for R of the following form: Rr Ar k where A and k are arbitrary constants. Substituting this exression in the differential euation for R(r) we obtain Ar k r kr 2 Ar k - k ( r k k +)r k kk ( +) mm ( +) Therefore, the constant k must satisfy the following relation: kk ( +) k 2 + k mm ( +) This euation gives us the following exression for k k - ± + 4m ( m + ) 2 - ± 2 m + ˆ Ë 2 2 Ï m Ô Ì or Ô Ó - m + The general solution for Rr is thus given by Rr Ar m + B r m + where A and B are arbitrary constants. The angle deendent art of the solution of Lalace's euation must satisfy the following euation a ˆ sin Ë Á + mm ( +)a( )sin The solutions of this euation are known as the Legendre olynomial P m ( cos ). The Legendre olynomials have the following roerties:. if m is even: P m ( x) P m (-x) 2. if m is odd: P m ( x) -P m (-x)

33 3. P m ( ) for all m 4. P n ( x)p ( m x )dx 2 Ú - 2m + d or P ( cos )P ( cos )sind 2 nm Ú - n m 2m + d nm Combining the solutions for Rr and a( ) we obtain the most general solution of Lalace's euation in a sherical symmetric system with azimuthal symmetry: Vr, Ë A m r m + B m r m + m ˆ P cos m Examle: Problem 3.8 The otential at the surface of a shere is given by V ( ) k cos( 3 ) where k is some constant. Find the otential inside and outside the shere, as well as the surface charge density s( ) on the shere. (Assume that there is no charge inside or outside of the shere.) The most general solution of Lalace's euation in sherical coordinates is Vr, Ë A m r m + B m r m + m ˆ P cos m First consider the region inside the shere (r < R). In this region B m since otherwise V( r, ) would blow u at r. Thus A m r m P m ( cos ) m Vr, The otential at r R is therefore eual to A m R m P m ( cos ) m VR, Using trigonometric relations we can rewrite cos( 3 ) as k cos( 3 ) cos( 3 ) 4 cos 3-3cos 8 5 P ( cos ) P( cos )

34 Substituting this exression in the euation for V( R, ) we obtain A m R m P m ( cos ) VR, 8k 5 P 3 cos m - 3k 5 P( cos ) This euation immediately shows that A m unless m orm 3. If m orm 3 then A A k R k R 3 The electrostatic otential inside the shere is therefore eual to Vr, 8k 5 r 3 R 3 P 3 ( cos ) - 3k 5 r R P( cos ) Now consider the region outsider the shere (r > R). In this region A m since otherwise Vr, would blow u at infinity. The solution of Lalace's euation in this region is therefore eual to Vr, The otential at r R is therefore eual to VR, B m B m m r m + P m cos R m + P ( m cos ) 8k 5 P cos 3 m - 3k 5 P( cos ) The euation immediately shows that B m excet when m orm 3. If m orm 3 then B kr2 B kr4 The electrostatic otential outside the shere is thus eual to Vr, - 3k 5 R 2 r 2 P cos + 8k 5 R 4 r 4 P 3 cos

35 The charge density on the shere can be obtained using the boundary conditions for the electric field at a boundary: E r R + - E r R - s( ) r ˆ Since E - V this boundary condition can be rewritten as e V - V - s r r R + r r R - e The first term on the left-hand side of this euation can be calculated using the electrostatic otential just obtained: V Á 6k r r R + Ë 5 R 2 r 3 P ( cos ) - 32k R 4 5 ˆ r 5 P 3 ( cos ) r R + ( - 32P ( cos )) 3 k 5R 6P cos In the same manner we obtain V Á - 3k r r R - Ë 5 Therefore, R P( cos ) + 24k 5 ˆ r 2 R 3 P 3 ( cos ) r R - ( + 24P ( cos )) 3 k 5R -3P cos V - V k r r R + r r R - 5R 9P cos The charge density on the shere is thus eual to ( - 56P ( cos )) -s ( ) 3 s( ) ke 5R -9P( cos ) + 56P ( cos ) 3 e Examle: Problem 3.9 Suose the otential V inside or outside the shere. Show that the charge density on the shere is given by where at the surface of a shere is secified, and there is no charge s( ) e 2R 2m + m 2 C m P m ( cos )

36 C m Ú V ( )P m ( cos )sind Most of the solution of this roblem is very similar to the solution of Problem 3.8. First consider the electrostatic otential inside the shere. The electrostatic otential in this region is given by and the boundary condition is VR, A m r m P m ( cos ) Vr, m m A m R m P m ( cos ) V ( ) The coefficients A m can be determined by multilying both sides of this euation by ( cos )sin and integrating with resect to between and : P n Thus V ( )P ( Ú cos )sind A n m R m Ú P ( m cos )P ( cos )sind n m 2 2n + A n Rn A m 2m + 2 R m Ú V ( )P m ( cos )sind In the region outside the shere the electrostatic otential is given by and the boundary condition is The coefficients B m are given by Vr, VR, B m m B m m r m + P m R m + P m cos cos V ( ) B m 2m + 2 R m + Ú V ( )P m ( cos )sind The charge density s( ) on the surface of the shere is eual to

37 Ï Ô V s( ) -e Ì Ó Ô r r R + - V r r R - Ô Ô Differentiating V( r, ) with resect to r in the region r > R we obtain V - ( m +) B m r r R + R m + 2 P m cos m Differentiating V( r, ) with resect to r in the region r < R we obtain The charge density is therefore eual to V ma r m R m - P m cos r R - m s( ) -e - ( m +) B m R P ( cos ) Ï È Ì m + 2 m Î - [ ma m R m - P m ( cos )] Ó m m Ï È e ( m +) B m Ì Ë R + ma m + 2 m Ó m Î Rm - ˆ P ( cos ) m Substituting the exressions for A m and B m into this euation we obtain Rm + Ï Ô È 2m + 2m + ˆ s( ) e Á ( m +) 2R m m 2R m R m - Ô Ì C Ë m P m ( cos ) Ó Ô m Î Ô e 2R Ï Ì Ó [ ( 2m +) 2 C m P m ( cos )] m where C m Ú V ( )P m ( cos )sind Examle: Problem 3.23 Solve Lalace's euation by searation of variables in cylindrical coordinates, assuming there is no deendence on z (cylindrical symmetry). Make sure that you find all solutions to the radial euation. Does your result accommodate the case of an infinite line charge?

38 For a system with cylindrical symmetry the electrostatic otential does not deend on z. This immediately imlies that V / z. Under this assumtion Lalace's euation reads r r r V ˆ Ë r + r 2 2 V f 2 Consider as a ossible solution of V: Vr,f Rr a( f ) Substituting this solution into Lalace's euation we obtain a f r r r R r ˆ Ë Á r + Rr 2 a f r 2 f 2 Multilying each term in this euation by r 2 and dividing by Rr a( f ) we obtain r Rr r r R r ˆ Ë Á r + 2 a f a f f 2 The first term in this euation deends only on r while the second term in this euation deends only on f. This euation can therefore be only valid for every r and every f if each term is eual to a constant. Thus we reuire that Rr r r R r ˆ Ë Á r constant g and a f 2 a f f 2 -g First consider the case in which g -m 2 <. The differential euation for a( f ) can be rewritten as 2 a( f ) f 2 - m 2 a f The most general solution of this differential solution is

39 a m ( f ) C m e mf + D m e -mf However, in cylindrical coordinates we reuire that any solution for a given f is eual to the solution for f + 2. Obviously this condition is not satisfied for this solution, and we conclude that g m 2. The differential euation for a f 2 a f f 2 + m 2 a f The most general solution of this differential solution is can be rewritten as a m ( f ) C m cos( mf ) + D m sin( mf ) a m ( f + 2 ) reuires that m is an integer. Now consider the radial The condition that a m f function Rr. We will first consider the case in which g m 2 >. Consider the following solution for Rr : Rr Ar k Substituting this solution into the revious differential euation we obtain r Ar k r r ( r Ar k ˆ Ë ) Ar k - r r kark - ( ) Ar k - r kark Ar k - k 2 Ar k - k 2 m 2 Therefore, the constant k can take on the following two values: k + m k - -m The most general solution for Rr under the assumtion that m 2 > is therefore Rr A m r m + B m r m Now consider the solutions for Rr when m 2. In this case we reuire that or r r R r ˆ Ë Á r

40 r Rr constant a r This euation can be rewritten as R( r) a r r If a then the solution of this differential euation is Rr b constant If a π then the solution of this differential euation is Rr a ln( r ) + b Combining the solutions obtained for m 2 with the solutions obtained for m 2 > we conclude that the most general solution for Rr is given by Rr a ln r È Î + b + A m r m + B m r m m Therefore, the most general solution of Lalace's euation for a system with cylindrical symmetry is Vr,f a ln r È Î Ë + b + A m r m + B m r m m ˆ C m cos mf ( + D m sin( mf )) Examle: Problem 3.25 A charge density s a sin( 5f ) is glued over the surface of an infinite cylinder of radius R. Find the otential inside and outside the cylinder. The electrostatic otential can be obtained using the general solution of Lalace's euation for a system with cylindrical symmetry obtained in Problem In the region inside the cylinder the coefficient B m must be eual to zero since otherwise Vr,f For the same reason a. Thus would blow u at r

41 b in, + r m C in,m cos( mf ) + D in,m sin( mf ) Vr,f m [ ] In the region outside the cylinder the coefficients A m must be eual to zero since otherwise Vr,f would blow u at infinity. For the same reason a. Thus Vr,f b out, + È Î r m m ( C out, m cos( mf ) + D out, m sin( mf )) Since Vr,f must aroach when r aroaches infinity, we must also reuire that b out, is eual to. The charge density on the surface of the cylinder is eual to V -e s f È Î r r R + - V r r R - Differentiating V( r,f ) in the region r > R and setting r R we obtain V È - m r r R + Î m R m + ( C out, m cos( mf ) + D out, m sin( mf )) Differentiating V( r,f ) in the region r < R and setting r R we obtain ( + D in, m sin( mf )) m [ ] V mr m - C r in,m cos mf r R - The charge density on the surface of the cylinder is therefore eual to È e mr m - C in,m + m s f m Î Ë ˆ R m + C out, m cos mf + mr m - D in,m + m Ë ˆ R m + D out, m sin ( mf ) Since the charge density is roortional to sin( 5f ) we can conclude immediately that C in, m C out, m for all m and that D in, m D out, m for all m excet m 5. Therefore This reuires that e 5R 4 D in,5 + 5 R 6 D out,5 s f Ë ˆ sin 5f 5R 4 D in,5 + 5 R 6 D out,5 a sin ( 5f ) a e - 4 -

42 A second relation between D in,5 and D out,5 can be obtained using the condition that the electrostatic otential is continuous at any boundary. This reuires that V in ( R,f ) b in, + R 5 D in,5 sin 5f V out ( R,f ) D out,5 R 5 sin( 5f ) Thus b in, and D out,5 R D in,5 We now have two euations with two unknown, D in,5 and D out,5, which can be solved with the following result: D in,5 a e and D out,5 a R 6 e The electrostatic otential inside the cylinder is thus eual to V in ( r,f ) r 5 D in,5 sin 5f R 4 a r 5 e R 4 sin 5f The electrostatic otential outside the cylinder is thus eual to V out ( r,f ) D out,5 r 5 sin( 5f ) a R 6 e r 5 sin( 5f ) Examle: Problem 3.37 A conducting shere of radius a, at otential V, is surrounded by a thin concentric sherical shell of radius b, over which someone has glued a surface charge s( ) s cos

43 where s is a constant. a) Find the electrostatic otential in each region: i) r > b ii) a < r < b b) Find the induced surface charge s( ) on the conductor. c) What is the total charge of the system? Check that your answer is consistent with the behavior of V at large r. a) The system has sherical symmetry and we can therefore use the most general solution of Lalace's euation in sherical coordinates: Vr, Ë A m r m + B m r m + m ˆ P cos m In the region inside the shere B m since otherwise V( r, ) would blow u at r. Therefore A m r m P m ( cos ) Vr, The boundary condition for V( r, ) is that it is eual to V at r a: Va, m A m a m P m ( cos ) m V V P cos This immediately shows that A m for all m excet m : A V The electrostatic otential inside the shere is thus given by V r <a ( r, ) V which should not come as a surrise. In the region outside the shell A m since otherwise Vr, Thus V r >b ( r, ) B out, m r m + P m ( cos ) m would blow u at infinity. In the region between the shere and the shell the most general solution for Vr, is given by

44 Á Ë V a <r <b ( r, ) A in,m r m + B in,m r m + m ˆ P m cos The boundary condition for V a <r <b ( r, ) at r a is V a <r <b A in,m a m + B in,m a, ˆ Á P Ë ( m cos ) V V P cos m a m + This euation can only be satisfied if A in, m a m + B in,m a m + if m > A in, + B in, a V if m The reuirement that the electrostatic otential is continuous at r b reuires that or B in,m b m + ˆ Á A in,m b m + P Ë ( m cos ) m B out, m b m + P m ( cos ) m This condition can be rewritten as A in, m b m + B in,m b m + B out, m b m + 2m + B out, m - B in,m A in,m b The other boundary condition for the electrostatic otential at r b is that it must roduce the charge distribution given in the roblem. This reuires that V -e s f È Î r r b + - V r r b - e m + ( B b m + 2 out, m - B in,m ) + ma in,m b m - ˆ Ë P ( cos ) m m s cos s P ( cos ) This condition is satisfied if m + ( b m + 2 B out, m - B in,m ) + ma in,m b m - if m π

45 2 ( B out, - B in, ) + A in, s if m e b 3 Substituting the relation between the various coefficients obtained by alying the continuity condition we obtain m + b m + 2 A in, m b 2m + + ma in,m b m - ( 2m +)A in,m b m - if m π 2 b 3 A in, b 3 + A in, 3A in, s e if m These euations show that A in, m if m π A in, s 3e if m Using these values for A in, m we can show that B out, m - B in,m if m π B out, - B in, s 3e b 3 if m The boundary condition for V at r a shows that B in, m -A in,m a 2m + if m 2 B in, -A in, a 3 - s 3e a 3 if m B in, av ( - A in, ) av if m These values for B in, m immediately fix the values for B out, m : B out, m B in,m if m 2 B out, s b 3 + B 3e in, s ( b 3 - a 3 ) if m 3e B out, B in, av if m

46 The otential in the region outside the shell is therefore eual to V r >b av r, r P ( cos ) + s 3e ( r 2 b 3 - a 3 )P cos The otential in the region between the shere and the shell is eual to V a <r <b av r, r P + s cos r - a 3 ˆ Á 3e Ë r 2 P cos b) The charge density on the surface of the shere can be found by calculating the sloe of the electrostatic otential at this surface: È V s( ) -e Î r r a + - V r r a - -e È -V a + s cos e V Î e a - s cos c) The total charge on the shere is eual to 2 Q a Ú Ú s ( ) a 2 sin d df 2a 2 Ï 2 e V a - s Ì Ú cos sin d Ó 4 ae V The total charge on the shell is eual to zero. Therefore the total charge of the system is eual to Q total 4 ae V The electrostatic otential at large distances will therefore be aroximately eual to V Q total r 4 ae V av r r This is eual to limit of the exact electrostatic otential when r Æ Multiole Exansions to Consider a given charge distribution r. The otential at a oint P (see Figure 3.) is eual VP Ú Volume r d dt

47 where d is the distance between P and a infinitesimal segment of the charge distribution. Figure 3. shows that d can be written as a function of r, r' and : d 2 r 2 + r' 2-2rr 'cos r 2 + r' ˆ Á Ë Ë r 2-2 r' ˆ Ë r cos ˆ y r' d r P x This euation can be rewritten as Figure 3.. Charge distribution r. d r + r' ˆ Á Ë Ë r 2-2 r' ˆ Ë r cos ˆ At large distances from the charge distribution r >> r' and conseuently r'/r <<. Using the following exansion for / ( + x): we can rewrite /d as - 2 x x2-5 + x 6 x d ª Ï Ô r - r' ˆ r' ˆ Ì Á 2 Ë r Ë Ë r Ó Ô ˆ - 2cos + 3 r' ˆ 8 Ë r 2 r' ˆ Á Ë Ë r ˆ - 2cos 2 Ô -... Ô Ï Ô r + r' ˆ Ë r cos + r' ˆ Ì Ó Ô Ë r cos2 - ˆ Ë 2 Ô +... Ô r n r' ˆ Ë r n P n cos Using this exansion of /d we can rewrite the electrostatic otential at P as

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