and its sign is immaterial.From the definition of capacitance,we have C= 0-￡oA (parallel plate). (5.2.4) |△Vd Note that C depends only on the geometric factors 4 and d.The capacitance C increases linearly with the area 4 since for a given potential difference Al,a bigger plate can hold more charge.On the other hand,C is inversely proportional to d,the distance of separation because the smaller the value of d,the smaller the potential difference Al for a fixed O. Example 5.2:Cylindrical Capacitor Consider next a solid cylindrical conductor of radius a surrounded by a coaxial cylindrical shell of inner radius b,as shown in Figure 5.2.3.The length of both cylinders is L and we take this length to be much larger than b-a,the separation of the cylinders, so that edge effects can be neglected.The capacitor is charged so that the inner cylinder has charge +O while the outer shell has a charge-O.What is the capacitance? (a) (b) Figure 5.2.3 (a)A cylindrical capacitor.(b)End view of the capacitor.The electric field is non-vanishing only in the region a <r<b. Solution: To calculate the capacitance,we first compute the electric field everywhere.Due to the cylindrical symmetry of the system,we choose our Gaussian surface to be a coaxial cylinder with length <L and radius r where a<r<b.Using Gauss's law,we have ∯EdA=EA=E(2r)=2L E=- (5.2.5) 2πer where =/L is the charge per unit length.Notice that the electric field is non- vanishing only in the region a<r<b.For r<a,the enclosed charge is =0 because in electrostatic equilibrium any charge in a conductor must reside on its surface.Similarly, for rb,the enclosed charge is ene==0 since the Gaussian surface encloses equal but opposite charges from both conductors.The potential difference is given by 5-6
5-6 and its sign is immaterial. From the definition of capacitance, we have C = Q | !V | = " 0A d (parallelplate) . (5.2.4) Note that C depends only on the geometric factors A and d. The capacitance C increases linearly with the area A since for a given potential difference !V , a bigger plate can hold more charge. On the other hand, C is inversely proportional to d, the distance of separation because the smaller the value of d, the smaller the potential difference | !V | for a fixed Q. Example 5.2: Cylindrical Capacitor Consider next a solid cylindrical conductor of radius a surrounded by a coaxial cylindrical shell of inner radius b, as shown in Figure 5.2.3. The length of both cylinders is L and we take this length to be much larger than b−a, the separation of the cylinders, so that edge effects can be neglected. The capacitor is charged so that the inner cylinder has charge +Q while the outer shell has a charge –Q. What is the capacitance? (a) (b) Figure 5.2.3 (a) A cylindrical capacitor. (b) End view of the capacitor. The electric field is non-vanishing only in the region a < r < b. Solution: To calculate the capacitance, we first compute the electric field everywhere. Due to the cylindrical symmetry of the system, we choose our Gaussian surface to be a coaxial cylinder with length ! < L and radius r where a < r < b . Using Gauss’s law, we have E !" ! d A !" S #"" = EA = E( 2#r! ) = $! % 0 & E = $ 2#% 0 r , (5.2.5) where ! = Q / L is the charge per unit length. Notice that the electric field is nonvanishing only in the region a < r < b . For r < a , the enclosed charge is enc q = 0 because in electrostatic equilibrium any charge in a conductor must reside on its surface. Similarly, for r > b , the enclosed charge is enc q = !! " !! = 0 since the Gaussian surface encloses equal but opposite charges from both conductors. The potential difference is given by
(5.2.6) where we have chosen the integration path to be along the direction of the electric field lines.As expected,the outer conductor with negative charge has a lower potential.The capacitance is then Q AL 2πeL C= (5.2.7) |△'|2ln(b/a)/2re。ln(b/a) Once again,we see that the capacitance C depends only on the length L,and the radii a and b. Example 5.3:Spherical Capacitor As a third example,let's consider a spherical capacitor which consists of two concentric spherical shells of radii a and b,as shown in Figure 5.2.4.The inner shell has a charge +O uniformly distributed over its surface,and the outer shell an equal but opposite charge-O.What is the capacitance of this configuration? Gaussian surface 0 (a) (b) Figure 5.2.4 (a)spherical capacitor with two concentric spherical shells of radii a and b. (b)Gaussian surface for calculating the electric field. Solution:The electric field is non-vanishing only in the region a<r<b.Using Gauss's law,we obtain ∯E,=E4-E4红)=号 (5.2.8) The radial component of the electric field is then E.= 1 (5.2.9) 4πe。r2 5-7
5-7 0 0 ln 2 2 b b a r a b a dr b V V V E dr r a ! ! "# "# $ % & = ' = ' = ' = ' ( ) * + , , , (5.2.6) where we have chosen the integration path to be along the direction of the electric field lines. As expected, the outer conductor with negative charge has a lower potential. The capacitance is then 0 0 2 | | ln( / )/ 2 ln( / ) Q L L C V b a b a # !" # !" = = = $ . (5.2.7) Once again, we see that the capacitance C depends only on the length L, and the radii a and b. Example 5.3: Spherical Capacitor As a third example, let’s consider a spherical capacitor which consists of two concentric spherical shells of radii a and b, as shown in Figure 5.2.4. The inner shell has a charge +Q uniformly distributed over its surface, and the outer shell an equal but opposite charge –Q. What is the capacitance of this configuration? (a) (b) Figure 5.2.4 (a) spherical capacitor with two concentric spherical shells of radii a and b. (b) Gaussian surface for calculating the electric field. Solution: The electric field is non-vanishing only in the region a < r < b . Using Gauss’s law, we obtain E !" ! dA !" S #"" = Er A = Er (4# r 2 ) = Q $ 0 . (5.2.8) The radial component of the electric field is then 2 1 4 r o Q E !" r = . (5.2.9)
Therefore,the potential difference between the two conducting shells is: (5.2.10) which yields for the capacitance Q =4πe0 ab C (5.2.11) I△VI b-a The capacitance C depends only on the radii a and b. An "isolated"conductor (with the second conductor placed at infinity)also has a capacitance.In the limit where b→∞，the above equation becomes ab lim C=lim b-a lim4πeo 4πeoa (5.2.12) Thus,for a single isolated spherical conductor of radius R,the capacitance is C=4πeoR. (5.2.13) The above expression can also be obtained by noting that a conducting sphere of radius R with a charge O uniformly distributed over its surface has V=O/4eR,where infinity is the reference point at zero potential,V()=0.Using our definition for capacitance, C=0 Q -=4πeR. (5.2.14) I△V|Q/4πeoR As expected,the capacitance of an isolated charged sphere only depends on the radius R. 5.3 Storing Energy in a Capacitor A capacitor can be charged by connecting the plates to the terminals of a battery,which are maintained at a potential difference AV called the terminal voltage. 5-8
5-8 Therefore, the potential difference between the two conducting shells is: 2 0 0 0 1 1 4 4 4 b b b a r a a Q dr Q Q b a V V V E dr !" r !" a b !" ab $ % $ # % & = # = # = # = # ' # ( = # ' ( ) * ) * + + , (5.2.10) which yields for the capacitance 0 4 | | Q ab C V b a !" # $ = = % & ' ) ( * . (5.2.11) The capacitance C depends only on the radii a and b. An “isolated” conductor (with the second conductor placed at infinity) also has a capacitance. In the limit where b " ! , the above equation becomes 0 0 0 lim lim 4 lim 4 4 1 b b b ab a C a b a a b !" !" !" #$ #$ #$ % & = ' ( = = * ) + % & ' ) ( * + . (5.2.12) Thus, for a single isolated spherical conductor of radius R, the capacitance is 0 C = 4!" R . (5.2.13) The above expression can also be obtained by noting that a conducting sphere of radius R with a charge Q uniformly distributed over its surface has 0 V = Q / 4!" R , where infinity is the reference point at zero potential, V (!) = 0 . Using our definition for capacitance, 0 0 4 | | / 4 Q Q C R V Q R !" !" = = = # . (5.2.14) As expected, the capacitance of an isolated charged sphere only depends on the radius R. 5.3 Storing Energy in a Capacitor A capacitor can be charged by connecting the plates to the terminals of a battery, which are maintained at a potential difference !V called the terminal voltage
MIT Figure 5.3.1 Charging a capacitor. The connection results in sharing the charges between the terminals and the plates.For example,the plate that is connected to the (positive)negative terminal will acquire some (positive)negative charge.The sharing causes a momentary reduction of charges on the terminals,and a decrease in the terminal voltage.Chemical reactions are then triggered to transfer more charge from one terminal to the other to compensate for the loss of charge to the capacitor plates,and maintain the terminal voltage at its initial level.The battery could thus be thought of as a charge pump that brings a charge o from one plate to the other. As discussed in the introduction,capacitors can be used to stored electrical energy.The amount of energy stored is equal to the work done to charge it.During the charging process,the battery does work to remove charges from one plate and deposit them onto the other. +十+++++++++++ Figure 5.3.1 Work is done by an external agent in bringing +dg from the negative plate and depositing the charge on the positive plate. Let the capacitor be initially uncharged.In each plate of the capacitor,there are many negative and positive charges,but the number of negative charges balances the number of positive charges,so that there is no net charge,and therefore no electric field between the plates.We have a magic bucket and a set of stairs from the bottom plate to the top plate (Figure 5.3.1).We show a movie of what is essentially this process in Section 5.5.2 below. 5-9
5-9 Figure 5.3.1 Charging a capacitor. The connection results in sharing the charges between the terminals and the plates. For example, the plate that is connected to the (positive) negative terminal will acquire some (positive) negative charge. The sharing causes a momentary reduction of charges on the terminals, and a decrease in the terminal voltage. Chemical reactions are then triggered to transfer more charge from one terminal to the other to compensate for the loss of charge to the capacitor plates, and maintain the terminal voltage at its initial level. The battery could thus be thought of as a charge pump that brings a charge Q from one plate to the other. As discussed in the introduction, capacitors can be used to stored electrical energy. The amount of energy stored is equal to the work done to charge it. During the charging process, the battery does work to remove charges from one plate and deposit them onto the other. Figure 5.3.1 Work is done by an external agent in bringing +dq from the negative plate and depositing the charge on the positive plate. Let the capacitor be initially uncharged. In each plate of the capacitor, there are many negative and positive charges, but the number of negative charges balances the number of positive charges, so that there is no net charge, and therefore no electric field between the plates. We have a magic bucket and a set of stairs from the bottom plate to the top plate (Figure 5.3.1). We show a movie of what is essentially this process in Section 5.5.2 below
We start out at the bottom plate,fill our magic bucket with a charge +dg,carry the bucket up the stairs and dump the contents of the bucket on the top plate,charging it up positive to charge +dg.However,in doing so,the bottom plate is now charged to -dg. Having emptied the bucket of charge,we now descend the stairs,get another bucketful of charge +dg,go back up the stairs and dump that charge on the top plate.We then repeat this process over and over.In this way we build up charge on the capacitor,and create electric field where there was none initially. Suppose the amount of charge on the top plate at some instant is +g,and the potential difference between the two plates is Al=g/C.To dump another bucket of charge +dg on the top plate,the amount of work done to overcome electrical repulsion is dw=AVdg.If at the end of the charging process,the charge on the top plate is + then the total amount of work done in this process is m-1ar=是-是 (5.3.1) This is equal to the electrical potential energy Ue of the system: U,-1-QIAYHCIAPP 2C2 (5.3.2) 5.3.1 Energy Density of the Electric Field One can think of the energy stored in the capacitor as being stored in the electric field itself.In the case of a parallel-plate capacitor,with C=4/d and AV =Ed,we have U.CIAP() (5.3.3) Because the quantity Ad represents the volume between the plates,we can define the electric energy density as u E UE= (5.3.4) Volume 2 The energy density u is proportional to the square of the electric field.Alternatively,one may obtain the energy stored in the capacitor from the point of view of external work. Because the plates are oppositely charged,force must be applied to maintain a constant separation between them.From Eq.(3.4.7),we see that a small patch of charge Ag=(A4)experiences an attractive force AF=(AA)/2.If the total area of the plate is A,then an external agent must exert a force F=2A/2 to pull the two plates 5-10
5-10 We start out at the bottom plate, fill our magic bucket with a charge +dq , carry the bucket up the stairs and dump the contents of the bucket on the top plate, charging it up positive to charge +dq . However, in doing so, the bottom plate is now charged to !dq . Having emptied the bucket of charge, we now descend the stairs, get another bucketful of charge +dq, go back up the stairs and dump that charge on the top plate. We then repeat this process over and over. In this way we build up charge on the capacitor, and create electric field where there was none initially. Suppose the amount of charge on the top plate at some instant is +q , and the potential difference between the two plates is | !V |= q /C . To dump another bucket of charge +dq on the top plate, the amount of work done to overcome electrical repulsion is dW =| !V | dq . If at the end of the charging process, the charge on the top plate is +Q , then the total amount of work done in this process is 2 0 0 1 | | 2 Q Q q Q W dq V dq C C = ! = = " " . (5.3.1) This is equal to the electrical potential energy UE of the system: 2 1 1 1 2 | | | | 2 2 2 E Q U Q V C V C = = ! = ! . (5.3.2) 5.3.1 Energy Density of the Electric Field One can think of the energy stored in the capacitor as being stored in the electric field itself. In the case of a parallel-plate capacitor, with 0 C = ! A/ d and | !V |= Ed , we have UE = 1 2 C | !V | 2 = 1 2 " 0A d (Ed) 2 = 1 2 " 0E2 (Ad) . (5.3.3) Because the quantity Ad represents the volume between the plates, we can define the electric energy density as 2 0 1 Volume 2 E E U u = = ! E . (5.3.4) The energy density E u is proportional to the square of the electric field. Alternatively, one may obtain the energy stored in the capacitor from the point of view of external work. Because the plates are oppositely charged, force must be applied to maintain a constant separation between them. From Eq. (3.4.7), we see that a small patch of charge "q = ! ("A) experiences an attractive force 2 0 #F = ! (#A)/ 2" . If the total area of the plate is A, then an external agent must exert a force 2 ext 0 F = ! A/ 2" to pull the two plates