# Chapter 2. Electrostatics

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1 Chapte. Electostatics.. The Electostatic Field To calculate the foce exeted by some electic chages,,, 3,... (the souce chages) on anothe chage Q (the test chage) we can use the pinciple of supeposition. This pinciple states that the inteaction between any two chages is completely unaffected by the pesence of othe chages. The foce exeted on Q by,, and 3 (see Figue.) is theefoe eual to the vecto sum of the foce F exeted by on Q, the foce F exeted by on Q, and the foce F 3 exeted by 3 on Q. 3 F Ftot F3 Q F Figue.. Supeposition of foces. The foce exeted by a chaged paticle on anothe chaged paticle depends on thei sepaation distance, on thei velocities and on thei acceleations. In this Chapte we will conside the special case in which the souce chages ae stationay. The electic field poduced by stationay souce chages is called and electostatic field. The electic field at a paticula point is a vecto whose magnitude is popotional to the total foce acting on a test chage located at that point, and whose diection is eual to the diection of - -

2 the foce acting on a positive test chage. The electic field E, geneated by a collection of souce chages, is defined as E = whee F is the total electic foce exeted by the souce chages on the test chage Q. It is assumed that the test chage Q is small and theefoe does not change the distibution of the souce chages. The total foce exeted by the souce chages on the test chage is eual to F = F + F + F = Ê Á Ë Q F Q ˆ + Q ˆ + Q 3 ˆ 3 The electic field geneated by the souce chages is thus eual to E = F Q = n Â i = i ˆ i i ˆ = Q n i i = i In most applications the souce chages ae not discete, but ae distibuted continuously ove some egion. The following thee diffeent distibutions will be used in this couse: Â ˆ i. line chage l: the chage pe unit length.. suface chage s: the chage pe unit aea. 3. volume chage : the chage pe unit volume. To calculate the electic field at a point P geneated by these chage distibutions we have to eplace the summation ove the discete chages with an integation ove the continuous chage distibution:. fo a line chage: E ( P ) =. fo a suface chage: E ( P ) = 3. fo a volume chage: E ( P ) = ˆ ldl Line Suface Volume ˆ sda ˆ dt - -

3 Hee ˆ is the unit vecto fom a segment of the chage distibution to the point P at which we ae evaluating the electic field, and is the distance between this segment and point P. Example: Poblem. a) Find the electic field (magnitude and diection) a distance z above the midpoint between two eual chages a distance d apat. Check that you esult is consistent with what you would expect when z» d. b) Repeat pat a), only this time make he ight-hand chage - instead of +. Etot a) E El b) El P P Etot z E z d/ d/ d/ d/ Figue.. Poblem. a) Figue.a shows that the x components of the electic fields geneated by the two point chages cancel. The total electic field at P is eual to the sum of the z components of the electic fields geneated by the two point chages: E ( P ) = Ê 4 d + z ˆ Ë z 4 d + z z ˆ = p z Ê Ë 4 d + z ˆ z ˆ 3/ When z» d this euation becomes appoximately eual to E ( P p z z ˆ = z z ˆ - 3 -

4 which is the Coulomb field geneated by a point chage with chage. b) Fo the electic fields geneated by the point chages of the chage distibution shown in Figue.b the z components cancel. The net electic field is theefoe eual to E ( P ) = Ê 4 d + z ˆ Ë d 4 d + z x ˆ = d Ê Ë 4 d + z ˆ x ˆ 3/ Example: Poblem.5 Find the electic field a distance z above the cente of a cicula loop of adius which caies a unifom line chage l. dez de del P z Figue.3. Poblem.5. Each segment of the loop is located at the same distance fom P (see Figue.3). The magnitude of the electic field at P due to a segment of the ing of length dl is eual to de = ldl + z - 4 -

5 When we integate ove the whole ing, the hoizontal components of the electic field cancel. We theefoe only need to conside the vetical component of the electic field geneated by each segment: de z = z + z de = ldl z + z ( ) 3/ The total electic field geneated by the ing can be obtained by integating de z ove the whole ing: E = l z ( + z ) 3/ dl = 4pe Ring 0 z ( ( + z ) 3/ p )l = z ( + z ) 3/ Example: Poblem.7 Find the electic field a distance z fom the cente of a spheical suface of adius R, which caies a unifom suface chage density s. Teat the case z < R (inside) as well as z > R (outside). Expess you answe in tems of the total chage on the suface. P z-cos z cos sin Figue.4. Poblem.7. Conside a slice of the shell centeed on the z axis (see Figue.4). The pola angle of this slice is and its width is d. The aea da of this ing is da = ( p sin )d = p sind The total chage on this ing is eual to - 5 -

6 d = sda = sind whee is the total chage on the shell. The electic field poduced by this ing at P can be calculated using the solution of Poblem.5: de = z - cos 8p ( + z - z cos ) 3/ sind The total field at P can be found by integating de with espect to : E = p z - cos 8p = 0 + z - z cos ( ) 3/ sind = p z - cos 8p ( ) = 0 + z - z cos 8p ( ) 3/ dcos This integal can be solved using the following elation: - z - y ( + z - zy ) 3/ dy z - y + z - zy ( ) 3/ =- d dz + z - zy Substituting this expession into the integal we obtain: E = - d 8p dz + z - zy dy = - d - 8p dz + z - zy -z - = =- 8p d dz Ï ( + z ) - - z Ì Ó z Outside the shell, z > and conseuently the electic field is eual to E = - d ( + z ) - z - 8p dz z ( ) =- d dz z = z Inside the shell, z < and conseuently the electic field is eual to E = - d ( + z ) - - z 8p dz z ( ) =- d dz = 0-6 -

7 Thus the electic field of a chaged shell is zeo inside the shell. The electic field outside the shell is eual to the electic field of a point chage located at the cente of the shell... Divegence and Cul of Electostatic Fields The electic field can be gaphically epesented using field lines. The diection of the field lines indicates the diection in which a positive test chage moves when placed in this field. The density of field lines pe unit aea is popotional to the stength of the electic field. Field lines oiginate on positive chages and teminate on negative chages. Field lines can neve coss since if this would occu, the diection of the electic field at that paticula point would be undefined. Examples of field lines poduced by positive point chages ae shown in Figue.5. a) b) Figue.5. a) Electic field lines geneated by a positive point chage with chage. b) Electic field lines geneated by a positive point chage with chage. The flux of electic field lines though any suface is popotional to the numbe of field lines passing though that suface. Conside fo example a point chage located at the oigin. The electic flux F E though a sphee of adius, centeed on the oigin, is eual to F E = E da = Suface Ê ˆ ˆ Ë ( sinddfˆ ) = Suface Since the numbe of field lines geneated by the chage depends only on the magnitude of the chage, any abitaily shaped suface that encloses will intecept the same numbe of field lines. Theefoe the electic flux though any suface that encloses the chage is eual to /. Using the pinciple of supeposition we can extend ou conclusion easily to systems containing moe than one point chage: - 7 -

8 F E = E da = Â E i da = e i 0 Â Suface i Suface We thus conclude that fo an abitay suface and abitay chage distibution i Suface E da = Q enclosed whee Q enclosed is the total chage enclosed by the suface. This is called Gauss's law. Since this euation involves an integal it is also called Gauss's law in integal fom. Using the divegence theoem the electic flux F E can be ewitten as F E = E da = ( E )dt Suface Volume We can also ewite the enclosed chage Q encl in tems of the chage density : Gauss's law can thus be ewitten as Q enclosed = dt Volume ( E )dt = dt Volume Since we have not made any assumptions about the integation volume this euation must hold fo any volume. This euies that the integands ae eual: Volume E = This euation is called Gauss's law in diffeential fom. Gauss's law in diffeential fom can also be obtained diectly fom Coulomb's law fo a chage distibution ( '): E ( ') = Volume Dˆ D ( ) ' whee D = - '. The divegence of E is eual to ( )dt ' - 8 -

9 E ( ) = Ê Dˆ ˆ Á Ë D ( ')dt ' = 4pd 3 ( - ') 4pe ( ' )dt ' 0 = Volume ( ) which is Gauss's law in diffeential fom. Gauss's law in integal fom can be obtained by integating E ( ) ove the volume V: ( E ( ) )dt = E da =F E = Volume Suface Volume Volume ( ) dt = Q Enclosed ( ) Example: Poblem.4 If the electic field in some egion is given (in spheical coodinates) by the expession E ( ) = Aˆ + B sin cosf f ˆ whee A and B ae constants, what is the chage density? The chage density can be obtained fom the given electic field, using Gauss's law in diffeential fom: = E Ê = Á Ë Ê Ë ( ) = Á ( A ) + sin ( E ) + sin sine ( ) + Ê B sin cosf ˆ ˆ A f Ë = e 0 - e 0 sin f B sin f ( E f ) ˆ =... The cul of E Conside a chage distibution (). The electic field at a point P geneated by this chage distibution is eual to E ( ) = whee D = - '. The cul of E is eual to E ( ) = Dˆ D ( ) ( ') dt ' Ê Dˆ Á Ë D ( ) ˆ ( ') dt ' - 9 -

10 Howeve, ˆ / = 0 fo evey vecto and we thus conclude that E ( ) = 0... Applications of Gauss's law Although Gauss's law is always tue it is only a useful tool to calculate the electic field if the chage distibution is symmetic:. If the chage distibution has spheical symmety, then Gauss's law can be used with concentic sphees as Gaussian sufaces.. If the chage distibution has cylindical symmety, then Gauss's law can be used with coaxial cylindes as Gaussian sufaces. 3. If the chage distibution has plane symmety, then Gauss's law can be used with pill boxes as Gaussian sufaces. Example: Poblem. Use Gauss's law to find the electic field inside a unifomly chaged sphee (chage density ) of adius R. The chage distibution has spheical symmety and conseuently the Gaussian suface used to obtain the electic field will be a concentic sphee of adius. The electic flux though this suface is eual to F E = Suface E da = 4p E ( ) The chage enclosed by this Gaussian suface is eual to Q Enclosed = 4 3 p 3 Applying Gauss's law we obtain fo the electic field: - 0 -

11 E ( ) = Q Enclosed 4p = 4p 4 3 p 3 = 3 Example: Poblem.4 Find the electic field inside a sphee which caies a chage density popotional to the distance fom the oigin: = k, fo some constant k. The chage distibution has spheical symmety and we will theefoe use a concentic sphee of adius as a Gaussian suface. Since the electic field depends only on the distance, it is constant on the Gaussian suface. The electic flux though this suface is theefoe eual to F E = Suface E da = 4p E ( ) The chage enclosed by the Gaussian suface can be obtained by integating the chage distibution between ' = 0 and ' = : Applying Gauss's law we obtain: o Q Enclosed = ( ')dt = k' ( 4p' )d' = pk 4 Volume F E = 4p E ( ) = Q Enclosed = pk 4 Ê pk 4 ˆ Ë Á e E ( ) 0 = 4p = k 4 0 Example: Poblem.6 A long coaxial cable caies a unifom (positive) volume chage density on the inne cylinde (adius a), and unifom suface chage density on the oute cylindical shell (adius b). The suface chage is negative and of just the ight magnitude so that the cable as a whole is neutal. Find the electic field in each of the thee egions: () inside the inne cylinde ( < a), () between the cylindes (a < < b), (3) outside the cable (b < ). - -

12 The chage distibution has cylindical symmety and to apply Gauss's law we will use a cylindical Gaussian suface. Conside a cylinde of adius and length L. The electic field geneated by the cylindical chage distibution will be adially diected. As a conseuence, thee will be no electic flux going though the end caps of the cylinde (since hee E ^da ). The total electic flux though the cylinde is eual to F E = Suface E da = ple( ) The enclosed chage must be calculated sepaately fo each of the thee egions:. < a: Q Enclosed = p L. a < < b: Q Enclosed = pa L 3. b < : Q Enclosed = 0 Applying Gauss's law we find E ( ) = Q Enclosed pl Substituting the calculated Q encl fo the thee egions we obtain. < a: E ( ) = Q Enclosed = p L =. pl pl. a < < b: E ( ) = Q Enclosed = pa L = a pl pl 3. b < E( ) = Q Enclosed = 0 pl Example: Poblem.8 Two sphees, each of adius R and caying unifom chage densities of + and -, espectively, ae placed so that they patially ovelap (see Figue.6). Call the vecto fom the negative cente to the positive cente s. Show that the field in the egion of ovelap is constant and find its value. To calculate the total field geneated by this chage distibution we use the pinciple of supeposition. The electic field geneated by each sphee can be obtained using Gauss' law (see - -

13 Poblem.). Conside an abitay point in the ovelap egion of the two sphees (see Figue.7). The distance between this point and the cente of the negatively chaged sphee is -. The distance between this point and the cente of the positively chaged sphee is +. Figue.7 shows that the vecto sum of s and + is eual to -. Theefoe, =-s The total electic field at this point in the ovelap egion is the vecto sum of the field due to the positively chaged sphee and the field due to the negatively chaged sphee: E tot = ( 3e ) s Figue.6. Poblem.8. + s E+ E- Etot - Figue.7. Calculation of E tot

14 The minus sign in font of - shows that the electic field geneated by the negatively chaged sphee is diected opposite to -. Using the elation between + and - obtained fom Figue.7 we can ewite E tot as E tot = - 3 s which shows that the field in the ovelap egion is homogeneous and pointing in a diection opposite to s..3. The Electic Potential The euiement that the cul of the electic field is eual to zeo limits the numbe of vecto functions that can descibe the electic field. In addition, a theoem discussed in Chapte states that any vecto function whose cul is eual to zeo is the gadient of a scala function. The scala function whose gadient is the electic field is called the electic potential V and it is defined as E = - V Taking the line integal of V between point a and point b we obtain a b V dl = Vb ( ) -V( a) =- E dl Taking a to be the efeence point and defining the potential to be zeo thee, we obtain fo V(b) ( ) =- E dl Vb The choice of the efeence point a of the potential is abitay. Changing the efeence point of the potential amounts to adding a constant to the potential: V' b b a =- E dl a' a' ( ) =- E dl whee K is a constant, independent of b, and eual to a b b - E dl K = - E dl Howeve, since the gadient of a constant is eual to zeo a a' a a b = K + Vb ( ) - 4 -

15 E'= - V ' =- V = E Thus, the electic field geneated by V' is eual to the electic field geneated by V. The physical behavio of a system will depend only on the diffeence in electic potential and is theefoe independent of the choice of the efeence point. The most common choice of the efeence point in electostatic poblems is infinity and the coesponding value of the potential is usually taken to be eual to zeo: b ( ) =- E dl Vb The unit of the electical potential is the Volt (V, V = Nm/C). Example: Poblem.0 One of these is an impossible electostatic field. Which one? a) E = k[ ( xy)ˆ i + ( yz )ˆ j + ( 3xz )ˆ k ] b) E = k ( y )ˆ i + ( xy + z )ˆ j + ( yz )ˆ k [ ] Hee, k is a constant with the appopiate units. Fo the possible one, find the potential, using the oigin as you efeence point. Check you answe by computing V. a) The cul of this vecto function is eual to Ê E = ká Ë y 3xz Ê ká Ë x yz ( ) - ( ) - ( ) ( z yz ) ˆ ˆ Ê i + k Ë z xy ( ) - y xy ˆ = k -yˆ i - 3zˆ j - xˆ k ( ) x ( 3xz ) ˆ ˆ j + Since the cul of this vecto function is not eual to zeo, this vecto function can not descibe an electic field. b) The cul of this vecto function is eual to Ê E = ká Ë y yz z xy + z ˆ ˆ Ê i + k Ë ( ) - ( ) Ê ká x xy + z Ë ( ) - ( ) y y ˆ = 0 ( ) - ( ) z y x yz ˆ ˆ j

16 Since the cul of this vecto function is eual to zeo it can descibe an electic field. To calculate the electic potential V at an abitay point (x, y, z), using (0, 0, 0) as a efeence point, we have to evaluate the line integal of E between (0, 0, 0) and (x, y, z). Since the line integal of E is path independent we ae fee to choose the most convenient integation path. I will use the following integation path: 0, 0, 0 Æ x,0,0 Æ x, y,0 Æ x, y, z The fist segment of the integation path is along the x axis: and dl = dxˆ i E dl = ky dx = 0 since y = 0 along this path. Conseuently, the line integal of E along this segment of the integation path is eual to zeo. The second segment of the path is paallel to the y axis: and dl = dyˆ j E dl = k( xy + z )dy = kxydy since z = 0 along this path. The line integal of E along this segment of the integation path is eual to (x,y,0) y E dl = kxydy = kxy (x,0,0) The thid segment of the integation path is paallel to the z axis: and 0 dl = dzˆ k E dl = k( yz)dz The line integal of E along this segment of the integation path is eual to - 6 -

17 (x,y, z) z E dl = k( yz)dz = kyz (x, y,0) The electic potential at (x, y, z) is thus eual to (x,0,0) ( ) = - E dl Vx,y,z (0,0,0) 0 (x, y,0) - E dl - E dl = (x,0,0) = 0 - kxy - kyz =-kxy ( + yz ) The answe can be veified by calculating the gadient of V: (x, y, z) (x, y,0) V = V x ˆ i + V y ˆ j + V k z ˆ = -ky ( ˆ i + ( xy + z )ˆ j + ( yz )ˆ k ) = -E which is the opposite of the oiginal electic field E. The advantage of using the electic potential V instead of the electic field is that V is a scala function. The total electic potential geneated by a chage distibution can be found using the supeposition pinciple. This popety follows immediately fom the definition of V and the fact that the electic field satisfies the pinciple of supeposition. Since it follows that E = E + E + E b b b b V = - E dl = - E dl - E dl - E 3 dl -... =V + V + V This euation shows that the total potential at any point is the algebaic sum of the potentials at that point due to all the souce chages sepaately. This odinay sum of scalas is in geneal easie to evaluate then a vecto sum. Example: Poblem.46 Suppose the electic potential is given by the expession V( ) = A e -l fo all (A and l ae constants). Find the electic field E ( ), the chage density ( ), and the total chage Q

18 The electic field E ( ) can be immediately obtained fom the electic potential: Ê Á Ë E ( ) =- V( ) = - A e -l ˆ Ê ˆ = la e -l Á Ë + A e -l ˆ ˆ The chage density ( ) can be found using the electic field E ( ) and the following elation: E ( ) = ( ) This expession shows that [ ] ( ) = E ( ) Substituting the expession fo the electic field E ( ) we obtain fo the chage density ( ): È Ê ˆ ( ) = A Í Á ( + l )e -l Î Ë ˆ = È Ê = A ( + l )e -l ˆ ˆ Í Ë Á + Î ˆ ( + l -l ( )e ) = È = AÍ 4p ( + l )e -l d 3 Î ( ) - l e -l = A 4pd 3 ( ) - l e -l Í The total chage Q can be found by volume integation of ( ): È Q tot = ( )dt = A 4pd 3 ( ) - l e -l Í 4p d = 0 Î Volume The integal can be solved easily: = A 4pd 3 ( ) È d - l e -l d Î Í 0 0 = =- A l e -l d 0 È Î e -l d =- d e -l d 0 dl =- d Ê ˆ 0 dl Ë l = l The total chage is thus eual to - 8 -

19 Q tot = - A The chage distibution V( ) ( ) can be diectly used to obtained fom the electic potential [ ] =- [ V( )] = - V ( ) = E ( ) This euation can be ewitten as V ( ) =- ( ) and is known as Poisson's euation. In the egions whee ( ) = 0 this euation educes to Laplace's euation: V( ) = 0 The electic potential geneated by a discete chage distibution can be obtained using the pinciple of supeposition: V tot whee V i at the oigin will geneate an electic potential V i n Â ( ) = V i i = ( ) is the electic potential geneated by the point chage i. A point chage i located ( ) eual to V i ( ) = - ( ) i = ' d' i ( ) In geneal, point chage i will be located at position i and the electic potential geneated by this point chage at position is eual to V i ( ) = i - i The total electic potential geneated by the whole set of point chages is eual to V tot ( ) = n Â i = i - i - 9 -

20 To calculate the electic potential geneated by a continuous chage distibution we have to eplace the summation ove point chages with an integation ove the continuous chage distibution. Fo the thee chage distibutions we will be using in this couse we obtain:. line chage l : V tot ( ) = Line l - ' dl'. suface chage s : V tot ( ) = Suface s - ' da' 3. volume chage : V tot ( ) = Volume - ' dt ' Example: Poblem.5 Using the geneal expession fo V in tems of find the potential at a distance z above the cente of the chage distibutions of Figue.8. In each case, compute E = - V. Suppose that we changed the ight-hand chage in Figue.8a to -. What is then the potential at P? What field does this suggest? Compae you answe to Poblem.b, and explain caefully any discepancy. a) P b) P c) P z z z d l L s R Figue.8. Poblem.35. a) The electic potential at P geneated by the two point chages is eual to - 0 -

21 V = 4 d + z + 4 d + z = p 4 d + z The electic field geneated by the two point chages can be obtained by taking the gadient of the electic potential: E = - V = - Ê Á Á z pe Á 0 Ë ˆ k ˆ = 4 d + z pe 0 z Ê Ë 4 d + z ˆ ˆ 3/ k If we change the ight-hand chage to - then the total potential at P is eual to zeo. Howeve, this does not imply that the electic field at P is eual to zeo. In ou calculation we have assumed ight fom the stat that x = 0 and y = 0. Obviously, the potential at P will theefoe not show an x and y dependence. This howeve not necessaily indicates that the components of the electic field along the x and y diection ae zeo. This can be demonstated by calculating the geneal expession fo the electic potential of this chage distibution at an abitay point (x,y,z): Vx,y,z ( ) = Ê x + Ë d ˆ - + 4pe + y + z 0 x - = Ê Ë d ˆ + y + z = Ê x + - Ë d ˆ 4pe + y + z 0 x - Ê Ë d ˆ + y + z The vaious components of the electic field can be obtained by taking the gadient of this expession: E x ( x,y,z ) =- V x = E y ( x,y,z ) =- V y = Ê x + Ë d ˆ x + Ê Ê Ë d ˆ ˆ Á + y + z Ë Ê Ê x + Ë d ˆ Á Ë y ˆ + y + z 3/ - 3/ - Ê x - Ë d ˆ x - Ê Ê Ë d ˆ ˆ Á + y + z Ë Ê Ê x - Ë d ˆ Á Ë y ˆ + y + z 3/ 3/ - -

22 E z ( x,y,z ) =- V z = Ê Ê x + Ë d ˆ Á Ë z ˆ + y + z 3/ - Ê Ê x - Ë d ˆ Á Ë z ˆ + y + z 3/ The components of the electic field at P = (0, 0, z) can now be calculated easily: E x ( 0,0,z ) = E y ( 0,0,z ) = 0 E z ( 0,0,z ) = 0 d Ê Ë 4 d + z ˆ b) Conside a small segment of the od, centeed at position x and with length dx. The chage on this segment is eual to ldx. The potential geneated by this segment at P is eual to 3/ dv = ldx x + z The total potential geneated by the od at P can be obtained by integating dv between x = - L and x = L L ldx V = = l ln L + z + L -L x + z [ ( ) - ln( L + z - L )] The z component of the electic field at P can be obtained fom the potential V by calculating the z component of the gadient of V. We obtain E z ( x,y,z ) = - l [ ( )] = z ln ( L + z + L ) - ln L + z - L È = - l Í Í Í Í Î z L + z L + z + L - z L + z L + z - L = l L z L + z c) Conside a ing of adius and width d. The chage on this ing is eual to d = s[ p ( + d ) - p ] = p s d - -

23 The electic potential dv at P geneated by this ing is eual to dv = d + z = s d + z The total electic potential at P can be obtained by integating dv between = 0 and = R: [ ] V = s R d = s R + z - z 0 + z The z component of the electic field geneated by this chage distibution can be obtained by taking the gadient of V: E z = - s z [ R + z - z ] =- s È z R + z - Í Î Í Example: Poblem.5 Find the electic field a distance z above the cente of a cicula loop of adius, which caies a unifom line chage l. The total chage Q on the ing is eual to The total electic potential V at P is eual to Q = prl V = prl R + z = l R R + z The z component of the electic field at P can be obtained by calculating the gadient of V: E z = - l z R R + z = l Rz R + z ( ) 3/ This is the same answe we obtained in the beginning of this Chapte by taking the vecto sum of the segments of the ing. We have seen so fa that thee ae thee fundamental uantities of electostatics: - 3 -

24 . The chage density. The electic field E 3. The electic potential V If one of these uantities is known, the othes can be calculated: Known Ø E V ˆ dt V = dt E = E = E V =- E dl V =- V E = - V In geneal the chage density and the electic field E do not have to be continuous. Conside fo example an infinitesimal thin chage sheet with suface chage s. The elation between the electic field above and below the sheet can be obtained using Gauss's law. Conside a ectangula box of height e and aea A (see Figue.9). The electic flux though the suface of the box, in the limit e Æ 0, is eual to F E = E da = ( E^above - E^below )A Suface Eabove A e s Ebelow Figue.9. Electic field nea a chage sheet

25 E^,above - E^, below = s whee E^,above and E^,below ae the pependicula components of the electic field above and below the chage sheet. Using Gauss's law and the ectangula box shown in Figue.9 as integation volume we obtain Q encl = sa = E^, e above - E^, below 0 ( )A This euation shows that the electic field pependicula to the chage sheet is discontinuous at the bounday. The diffeence between the pependicula component of the electic field above and below the chage sheet is eual to The tangential component of the electic field is always continuous at any bounday. This can be demonstated by calculating the line integal of E aound a ectangula loop of length L and height e (see Figue.0). The line integal of E, in the limit e Æ 0, is eual to E dl = E,above dl + E,below dl = ( E,above - E,below )L Eabove e s L Ebelow Figue.0. Paallel field close to chage sheet. Since the line integal of E aound any closed loop is zeo we conclude that ( E,above - E,below )L = 0 o E,above = E,below These bounday conditions fo E can be combined into a single fomula: - 5 -

26 E above - E below = s ˆ n whee n ˆ is a unit vecto pependicula to the suface and pointing towads the above egion. The electic potential is continuous acoss any bounday. This is a diect esults of the definition of V in tems of the line integal of E : below V above -V below = E dl If the path shinks the line integal will appoach zeo, independent of whethe E is continuous o discontinuous. Thus above V above =V below Example: Poblem.30 a) Check that the esults of examples 4 and 5 of Giffiths ae consistent with the bounday conditions fo E. b) Use Gauss's law to find the field inside and outside a long hollow cylindical tube which caies a unifom suface chage s. Check that you esults ae consistent with the bounday conditions fo E. c) Check that the esult of example 7 of Giffiths is consistent with the bounday conditions fo V. a) Example 4 (Giffiths): The electic field geneated by an infinite plane caying a unifom suface chage s is diected pependicula to the sheet and has a magnitude eual to E above = s ˆ k E below = - s ˆ k Theefoe, E above - E below = s ˆ k = s sheet ˆ k which is in ageement with the bounday conditions fo E

27 Example 5 (Giffiths): The electic field geneated by the two chage sheets is diected pependicula to the sheets and has a magnitude eual to E I = 0 E II = s ˆ i E III = 0 The change in the stength of the electic field at the left sheet is eual to E II - E I = s i ˆ = s left ˆ i The change in the stength of the electic field at the ight sheet is eual to E III - E II = - s ˆ i = s ight ˆ i These elations show ageement with the bounday conditions fo E. b) Conside a Gaussian suface of length L and adius. As a esult of the symmety of the system, the electic field will be diected adially. The electic flux though this Gaussian suface is theefoe eual to the electic flux though its cuved suface which is eual to F E = E da = p L E( ) The chage enclosed by the Gaussian suface is eual to zeo when < R. Theefoe E( ) = 0 when < R. When > R the chage enclosed by the Gaussian suface is eual to Q enclosed = p R L s The electic field fo > R, obtained using Gauss' law, is eual to E ( ) = Q enclosed p L ˆ = s R ˆ The magnitude of the electic field just outside the cylinde, diected adially, is eual to - 7 -

28 E outside = E ( R ) = s ˆ The magnitude of the electic field just inside the cylinde is eual to Theefoe, E inside = 0 E outside - E inside = s ˆ which is consistent with the bounday conditions fo E. c) Example 7 (Giffiths): the electic potential just outside the chaged spheical shell is eual to V outside =V outside ( z = R) = R s R = s R The electic potential just inside the chaged spheical shell is eual to V inside =V inside ( z = R) = s R These two euations show that the electic potential is continuous at the bounday..4. Wok and Enegy in Electostatics Conside a point chage located at the oigin. A point chage is moved fom infinity to a point a distance fom the oigin. We will assume that the point chage emains fixed at the oigin when point chage is moved. The foce exeted by on is eual to F = E whee E is the electic field geneated by. In ode to move chage we will have to exet a foce opposite to F. Theefoe, the total wok that must be done to move fom infinity to is eual to - 8 -

29 ( ) W = - F dl =- E dl = V whee V ( ) is the electic potential geneated by at position. Using the euation of V fo a point chage, the wok W can be ewitten as W = This wok W is the wok necessay to assemble the system of two point chages and is also called the electostatic potential enegy of the system. The enegy of a system of moe than two point chages can be found in a simila manne using the supeposition pinciple. Fo example, fo a system consisting of thee point chages (see Figue.) the electostatic potential enegy is eual to W = Ï Ì Ó Figue.. System of thee point chages. In this euation we have added the electostatic enegies of each pai of point chages. The geneal expession of the electostatic potential enegy fo n point chages is W = n Â i = n Â j =i + i j ij The lowe limit of j (= i + ) insues that each pai of point chages is only counted once. The electostatic potential enegy can also be witten as - 9 -

30 W = È Í Í Î Í n Â i = n Â j = j πi i j = ij n Â i i = n Â j = j πi j ij = n Â i = i V i whee V i is the electostatic potential at the location of i due to all othe point chages. When the chage of the system is not distibuted as point chages, but athe as a continuous chage distibution, then the electostatic potential enegy of the system must be ewitten as W = Volume Vdt Fo continuous suface and line chages the electostatic potential enegy is eual to and W = W = Suface Line svda lvdl Howeve, we have aleady seen in this Chapte that, V, and E cay the same euivalent infomation. The chage density, fo example, is elated to the electic field E : = ( E ) Using this elation we can ewite the electostatic potential enegy as W = = ( E )Vdt = Volume VE da + Suface Volume ( ( VE ))dt - Volume ( E E )dt ( E ( V) )dt = whee we have used one of the poduct ules of vecto deivatives and the definition of E in tems of V. In deiving this expession we have not made any assumptions about the volume consideed. This expession is theefoe valid fo any volume. If we conside all space, then the contibution of the suface integal appoaches zeo since VE will appoach zeo faste than /. Thus the total electostatic potential enegy of the system is eual to Volume

31 W = Volume E dt Example: Poblem.45 A sphee of adius R caies a chage density enegy of the configuation. Check you answe by calculating it in at least two diffeent ways. ( ) = k (whee k is a constant). Find the Method : The fist method we will use to calculate the electostatic potential enegy of the chaged sphee uses the volume integal of E to calculate W. The electic field geneated by the chaged sphee can be obtained using Gauss's law. We will use a concentic sphee of adius as the Gaussian suface. Fist conside the case in which < R. The chage enclosed by the Gaussian suface can be obtained by volume integation of the chage distibution: Q Enclosed = ( )dt = 4p k d = p k 4 Sphee The electic flux though the Gaussian suface is eual to F E = 4p E ( ) Applying Gauss's law we find fo the electic field inside the sphee ( < R): E ( ) = 4p Q Enclosed = 0 4p p k 4 = k 4 The electic field outside the sphee ( > R) can also be obtained using Gauss's law: E ( ) = 4p Q Enclosed = 4p p k R 4 = k R4 4 The total electostatic enegy can be obtained fom the electic field: W = E R È k ( )dt = p Í Î 4e R È k R 4 d + p 0 0 Í Î 4 d = 0 = pk 8 All Space 0 R 6 d R 8 8 d R + pk = pk È 8 7 R7 + R 7 Í Î = pk R

32 Method : An altenative way calculate the electostatic potential enegy is to use the following elation: W = Volume Vdt The electostatic potential V can be obtained immediately fom the electic field E. To evaluate the volume integal of V we only need to know the electostatic potential V inside the chaged sphee: ( ) =- E dl V =- È R k R 4 Í 4 d + k d = R Î The electostatic potential enegy of the system is thus eual to k 4R 3-3 ( ) W = ( )V ( )dt = 4p Sphee R k k ( 4R 3-3 ) d = 4p 0 k Ê R 7 - ˆ Ë 7 R7 = p k 7 R 7 which is eual to the enegy calculated using method..5. Metallic Conductos In a metallic conducto one o moe electons pe atom ae fee to move aound though the mateial. Metallic conductos have the following electostatic popeties:. The electic field inside the conducto is eual to zeo. If thee would be an electic field inside the conducto, the fee chages would move and poduce an electic field of thei own opposite to the initial electic field. Fee chages will continue to flow until the cancellation of the initial field is complete.. The chage density inside a conducto is eual to zeo. This popety is a diect esult of popety. If the electic field inside a conducto is eual to zeo, then the electic flux though any abitay closed suface inside the conducto is eual to zeo. This immediately implies that the chage density inside the conducto is eual to zeo eveywhee (Gauss's law). 3. Any net chage of a conducto esides on the suface

33 Since the chage density inside a conducto is eual to zeo, any net chage can only eside on the suface. 4. The electostatic potential V is constant thoughout the conducto. Conside two abitay points a and b inside a conducto (see Figue.). The potential diffeence between a and b is eual to Vb ( ) -V( a) =- E dl a b a b Figue.. Potential inside metallic conducto. Since the electic field inside a conducto is eual to zeo, the line integal of E between a and b is eual to zeo. Thus o Vb ( ) -V( a) = 0 Vb ( ) =V( a) 5. The electic field is pependicula to the suface, just outside the conducto. If thee would be a tangential component of the electic field at the suface, then the suface chage would immediately flow aound the suface until it cancels this tangential component. Example: A spheical conducting shell a) Suppose we place a point chage at the cente of a neutal spheical conducting shell (see Figue.3). It will attact negative chage to the inne suface of the conducto. How much induced chage will accumulate hee? b) Find E and V as function of in the thee egions < a, a < < b, and > b

34 a b Figue.3. A spheical conducting shell. a) The electic field inside the conducting shell is eual to zeo (popety of conductos). Theefoe, the electic flux though any concentic spheical Gaussian suface of adius (a<<b) is eual to zeo. Howeve, accoding to Gauss's law this implies that the chage enclosed by this suface is eual to zeo. This can only be achieved if the chage accumulated on the inside of the conducting shell is eual to -. Since the conducting shell is neutal and any net chage must eside on the suface, the chage on the outside of the conducting shell must be eual to +. b) The electic field geneated by this system can be calculated using Gauss's law. In the thee diffeent egions the electic field is eual to E ( ) = fo b < E ( ) = 0 fo a < < b E ( ) = fo < a The electostatic potential V() can be obtained by calculating the line integal of E fom infinity to a point a distance fom the oigin. Taking the efeence point at infinity and setting the value of the electostatic potential to zeo thee we can calculate the electostatic potential. The line integal of E has to be evaluated fo each of the thee egions sepaately. Fo b < : ( ) =- E V ( ) dl =- ' d' =

35 Fo a < < b: ( ) =- E V ( ) dl =- b ' d' = b Fo < a: ( ) =- E V ( ) dl =- b ' d' - ' d' = È a b - Î Í a + Figue.4. Abitaily shaped conducto. In this example we have looked at a symmetic system but the geneal conclusions ae also valid fo an abitaily shaped conducto. Fo example, conside the conducto with a cavity shown in Figue.4. Conside also a Gaussian suface that completely suounds the cavity (see fo example the dashed line in Figue.4). Since the electic field inside the conducto is eual to zeo, the electic flux though the Gaussian suface is eual to zeo. Gauss's law immediately implies that the chage enclosed by the suface is eual to zeo. Theefoe, if thee is a chage inside the cavity thee will be an induced chage eual to - on the walls of the cavity. On the othe hand, if thee is no chage inside the cavity then thee will be no chage on the walls of the cavity. In this case, the electic field inside the cavity will be eual to zeo. This can be demonstated by assuming that the electic field inside the cavity is not eual to zeo. In this case, thee must be at least one field line inside the cavity. Since field lines oiginate on a positive chage and teminate on a negative chage, and since thee is no chage inside the cavity, this field line must stat and end on the cavity walls (see fo example Figue.5). Now conside a closed loop, which follows the field line inside the cavity and has an abitay shape inside the conducto (see Figue.5). The line integal of E inside the cavity is definitely not eual to zeo since the magnitude of E is not eual to zeo and since the path is defined such that E and dl ae paallel. Since the electic field inside the conducto is eual to zeo, the path

36 integal of E inside the conducto is eual to zeo. Theefoe, the path integal of E along the path indicated in Figue.5 is not eual to zeo if the magnitude of E is not eual to zeo inside the cavity. Howeve, the line integal of E along any closed path must be eual to zeo and conseuently the electic field inside the cavity must be eual to zeo. Figue.5. Field line in cavity. Example: Poblem.35 A metal sphee of adius R, caying chage, is suounded by a thick concentic metal shell (inne adius a, oute adius b, see Figue.6). The shell caies no net chage. a) Find the suface chage density s at R, at a, and at b. b) Find the potential at the cente of the sphee, using infinity as efeence. c) Now the oute suface is touched to a gounding wie, which lowes its potential to zeo (same as at infinity). How do you answes to a) and b) change? a) Since the net chage of a conducto esides on its suface, the chage of the metal sphee will eside its suface. The chage density on this suface will theefoe be eual to s R = 4p R As a esult of the chage on the metal sphee thee will be a chage eual to - induced on the inne suface of the metal shell. Its suface chage density will theefoe be eual to s a = - 4p a Since the metal shell is neutal thee will be a chage eual to + on the outside of the shell. The suface chage density on the outside of the shell will theefoe be eual to

37 s b = 4p b a b R Figue.6. Poblem.35. b) The potential at the cente of the metal sphee can be found by calculating the line integal of E between infinity and the cente. The electic field in the egions outside the sphee and shell can be found using Gauss's law. The electic field inside the shell and sphee is eual to zeo. Theefoe, 0 V cente =- E dl =- b d - R d = È a b - Î Í a + R c) When the outside of the shell is gounded, the chage density on the outside will become zeo. The chage density on the inside of the shell and on the metal sphee will emain the same. The electic potential at the cente of the system will also change as a esult of gounding the oute shell. Since the electic potential of the oute shell is zeo, we do not need to conside the line integal of E in the egion outside the shell to detemine the potential at the cente of the sphee. Thus 0 V cente =- E dl = - R È b d = a R - Î Í a Conside a conducto with suface chage s, placed in an extenal electic field. Because the electic field inside the conducto is zeo, the bounday conditions fo the electic field euie that the field immediately above the conducto is eual to

38 E above = s ˆ n Eothe Ep,above da Ep,below Figue.7. Patch of suface of conducto. This electic field will exet a foce on the suface chage. Conside a small, infinitely thin, patch of the suface with suface aea da (see Figue.7). The electic field diectly above and below the patch is eual to the vecto sum of the electic field geneated by the patch, the electic field geneated by the est of the conducto and the extenal electic field. The electic field geneated by the patch is eual to E p, above = s ˆ n E p, below =- s ˆ n The emaining field, E othe, is continuous acoss the patch, and conseuently the total electic field above and below the patch is eual to E above = E othe + s ˆ n E below = E othe - s ˆ n These two euations show that E othe is eual to

39 E othe = E above ( ) + E below In this case the electic field below the suface is eual to zeo and the electic field above the suface is diectly detemined by the bounday condition fo the electic field at the suface. Thus E othe = Ê s ˆ Á n ˆ + 0 = s n ˆ Ë Since the patch cannot exet a foce on itself, the electic foce exeted on it is entiely due to the electic field E othe. The chage on the patch is eual to s da. Theefoe, the foce exeted on the patch is eual to df = s da E othe = s da n ˆ The foce pe unit aea of the conducto is eual to f = df da = s n ˆ This euation can be ewitten in tems of the electic field just outside the conducto as f = ( e E) n ˆ = 0 E n ˆ This foce is diected outwads. It is called the adiation pessue..6. Capacitos Conside two conductos (see Figue.8), one with a chage eual to +Q and one with a chage eual to -Q. The potential diffeence between the two conductos is eual to DV = V + -V - = - E dl

40 Since the electic field E is popotional to the chage Q, the potential diffeence DV will also be popotional to Q. The constant of popotionality is called the capacitance C of the system and is defined as C = Q DV +Q -Q Figue.8. Two conductos. The capacitance C is detemined by the size, the shape, and the sepaation distance of the two conductos. The unit of capacitance is the faad (F). The capacitance of a system of conductos can in geneal be calculated by caying out the following steps:. Place a chage +Q on one of the conductos. Place a chage of -Q on the othe conducto (fo a two conducto system).. Calculate the electic field in the egion between the two conductos. 3. Use the electic field calculated in step to calculate the potential diffeence between the two conductos. 4. Apply the esult of pat 3 to calculate the capacitance: C = Q DV We will now discuss two examples in which we follow these steps to calculate the capacitance. Example: Example. (Giffiths) Find the capacitance of two concentic shells, with adii a and b. Place a chage +Q on the inne shell and a chage -Q on the oute shell. The electic field between the shells can be found using Gauss's law and is eual to

41 E ( ) = Q ˆ a < < b The potential diffeence between the oute shell and the inne shell is eual to Va ( ) -V( b) =- E The capacitance of this system is eual to b b Q ( ) dl = - a d = Q Ê a a - ˆ Ë b C = Q DV = Q Q Ê a - ˆ Ë b = ab b - a -Q +Q a b Figue.9. Example.. A system does not have to have two conductos in ode to have a capacitance. Conside fo example a single spheical shell of adius R. The capacitance of this system of conductos can be calculated by following the same steps as in Example. Fist of all, put a chage Q on the conducto. Gauss's law can be used to calculate the electic field geneated by this system with the following esult: E ( ) = Q ˆ Taking infinity as the efeence point we can calculate the electostatic potential on the suface of the shell: ( ) =- E VR R R Q ( ) dl = - d = Q R - 4 -

42 Theefoe, the capacitance of the shell is eual to C = Q DV = Q Q R = R Let us now conside a paallel-plate capacito. The wok euied to chage up the paallelplate capacito can be calculated in vaious ways: Method : Since we ae fee to chose the efeence point and efeence value of the potential we will chose it such that the potential of the positively chages plate is +DV / and the potential of the negatively chaged plate is -DV /. The enegy of this chage distibution is then eual to W = Vdt = È Í Q DV Î + -Q Ë ( ) Ê -DV ˆ = Q DV ( ) = C DV ( ) Method : Let us look at the chaging pocess in detail. Initially both conductos ae unchaged and DV = 0. At some intemediate step in the chaging pocess the chage on the positively chaged conducto will be eual to. The potential diffeence between the conductos will then be eual to DV = C To incease the chage on the positively chaged conducto by d we have to move this chage d acoss this potential diffeence DV. The wok euied to do this is eual to dw = DV d = C d Theefoe, the total wok euied to chage up the capacito fom = 0 to = Q is eual to Q W = C d = Q 0 C = C DV ( ) Example: Poblem.40. Suppose the plates of a paallel-plate capacito move close togethe by an infinitesimal distance e, as a esult of thei mutual attaction

43 a) Use euation (.45) of Giffiths to expess the amount of wok done by electostatic foces, in tems of the field E and the aea of the plates A. b) Use euation (.40) of Giffiths to expess the enegy lost by the field in this pocess. a) We will assume that the paallel-plate capacito is an ideal capacito with a homogeneous electic field E between the plates and no electic field outside the plates. The electostatic foce pe unit suface aea is eual to f = E ˆ n The total foce exeted on each plate is theefoe eual to F = Af = AE ˆ n As a esult of this foce, the plates of the paallel-plate capacito move close togethe by an infinitesimal distance e. The wok done by the electostatic foces duing this movement is eual to W = F dl = A e E b) The total enegy stoed in the electic field is eual to W = E dt In an ideal capacito the electic field is constant between the plates and conseuently we can easily evaluate the volume integal of E : W = A d E whee d is the distance between the plates. If the distance between the plates is educed, then the enegy stoed in the field will also be educed. A eduction in d of e will educe the enegy stoed by an amount DW eual to DW = A d E - A ( d - e ) E = A e E which is eual to the wok done by the electostatic foces on the capacito plates (see pat a)

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