Q 60A (parallel plate) (5.2.4) |△d Note that C depends only on the geometric factors 4 and d.The capacitance C increases linearly with the area A 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 AV for a fixed O. Interactive Simulation 5.1:Parallel-Plate Capacitor This simulation shown in Figure 5.2.3 illustrates the interaction of charged particles inside the two plates of a capacitor. 。9,生4。 Figure 5.2.3 Charged particles interacting inside the two plates of a capacitor. Each plate contains twelve charges interacting via Coulomb force,where one plate contains positive charges and the other contains negative charges.Because of their mutual repulsion,the particles in each plate are compelled to maximize the distance between one another,and thus spread themselves evenly around the outer edge of their enclosure.However,the particles in one plate are attracted to the particles in the other,so they attempt to minimize the distance between themselves and their oppositely charged correspondents.Thus,they distribute themselves along the surface of their bounding box closest to the other plate. 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 5.2.4.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? 5
0 parallel plate Q A C | V | d ε = = ∆ ( ) (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. Interactive Simulation 5.1: Parallel-Plate Capacitor This simulation shown in Figure 5.2.3 illustrates the interaction of charged particles inside the two plates of a capacitor. Figure 5.2.3 Charged particles interacting inside the two plates of a capacitor. Each plate contains twelve charges interacting via Coulomb force, where one plate contains positive charges and the other contains negative charges. Because of their mutual repulsion, the particles in each plate are compelled to maximize the distance between one another, and thus spread themselves evenly around the outer edge of their enclosure. However, the particles in one plate are attracted to the particles in the other, so they attempt to minimize the distance between themselves and their oppositely charged correspondents. Thus, they distribute themselves along the surface of their bounding box closest to the other plate. 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 5.2.4. 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? 5
(a) (b) Figure 5.2.4 (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.dA=EA=E(2r)=2→E (5.2.5) En 2π6J 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 dene=0 since any net charge in a conductor must reside on its surface.Similarly,for r>b,the enclosed charge is gne=-=0 since the Gaussian surface encloses equal but opposite charges from both conductors The potential difference is given by Ar=-=-ed=2 (5.2.6) π6oJar 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.This gives Q L 2π6，L (5.2.7) |△V|ln(b/a)/2πEo In(b/a) Once again,we see that the capacitance C depends only on the geometrical factors,L,a and b. 6
(a) (b) Figure 5.2.4 (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 A < L and radius r where a r < < b . Using Gauss’s law, we have ( ) 0 0 2 2 S d EA E r E r λ λ π ε πε ⋅ = = = ⇒ = ∫∫ E A A A JG JG w (5.2.5) where λ = 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 since any net charge in a conductor must reside on its surface. Similarly, for , the enclosed charge is enc q = 0 r > b enc q = λA A − λ = 0 since the Gaussian surface encloses equal but opposite charges from both conductors. The potential difference is given by 0 0 ln 2 2 b b a r a b a dr b V V V E dr r a λ λ πε πε ⎛ ⎞ ∆ = − = − = − = − ⎜ ⎟ ⎝ ⎠ ∫ ∫ (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. This gives 0 0 2 | | ln( / )/ 2 ln( / ) Q L L C V b a b λ πε λ πε = = = ∆ a (5.2.7) Once again, we see that the capacitance C depends only on the geometrical factors, L, a and b. 6
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 5.2.5.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 Figure 5.2.5 (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 ∯EdA=E4=E(4r）-号 (5.2.8) Or E,= 1 e (5.2.9) 4π8。r2 Therefore,the potential difference between the two conducting shells is: (5.2.10) which yields C= =4π0 ab (5.2.11) b-a Again,the capacitance C depends only on the physical dimensions,a and b. An "isolated"conductor (with the second conductor placed at infinity)also has a capacitance.In the limit where b->oo,the above equation becomes 7
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 5.2.5. 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? Figure 5.2.5 (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 ( ) 2 0 4 r r S Q d E A E π r ε ⋅ = = = ∫∫ E A JG JG w (5.2.8) or 2 1 4 r o Q E πε r = (5.2.9) 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 πε r a πε b πε ⎛ ⎞ ⎛ − ∆ = − = − = − = − ⎜ ⎟ − = − ⎜ ⎝ ⎠ ⎝ ∫ ∫ ab ⎞ ⎟ ⎠ (5.2.10) which yields 0 4 | | Q C V b πε ⎛ = = ⎜ ∆ ⎝ ⎠ ab a ⎞ ⎟ − (5.2.11) Again, the capacitance C depends only on the physical dimensions, a and b. An “isolated” conductor (with the second conductor placed at infinity) also has a capacitance. In the limit whereb → ∞ , the above equation becomes 7
ab limC=lim4πeo =1im4π8o 0 =4π8oa (5.2.12) b-oc b-a b Thus,for a single isolated spherical conductor of radius R,the capacitance is C=4π6R (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/4mR,using infinity as the reference point having zero potential,V(o)=0.This gives C= 0 =4π8R (5.2.14) |△V|Q/4πER As expected,the capacitance of an isolated charged sphere only depends on its geometry, namely,the radius R. 5.3 Capacitors in Electric Circuits 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. MIT △V 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 Ofrom one plate to the other. 8
0 0 lim lim 4 lim 4 4 1 b b b ab a C a b a a b π 0 ε πε →∞ →∞ →∞ ⎛ ⎞ = ⎜ ⎟ = ⎝ ⎠ − ⎛ ⎞ ⎜ ⎟ − ⎝ ⎠ = πε R (5.2.12) Thus, for a single isolated spherical conductor of radius R, the capacitance is 0 C = 4πε (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= / 4πε R , using infinity as the reference point having zero potential,V ( ) ∞ = 0 . This gives 0 0 4 | | / 4 Q Q C V Q R πε R πε = = = ∆ (5.2.14) As expected, the capacitance of an isolated charged sphere only depends on its geometry, namely, the radius R. 5.3 Capacitors in Electric Circuits 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. 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. 8
5.3.1 Parallel Connection Suppose we have two capacitors C with charge O and C2 with charge that are connected in parallel,as shown in Figure 5.3.2. IAKHAKH-AYI C Ceq-C1+C2 C 9 lrl ArI Figure 5.3.2 Capacitors in parallel and an equivalent capacitor. The left plates of both capacitors Ci and C2 are connected to the positive terminal of the battery and have the same electric potential as the positive terminal.Similarly,both right plates are negatively charged and have the same potential as the negative terminal.Thus, the potential difference Al is the same across each capacitor.This gives 品 6& (5.3.1) These two capacitors can be replaced by a single equivalent capacitor Cwith a total charge Osupplied by the battery.However,since O is shared by the two capacitors,we must have Q=g+Q2=CI△VI+C21△V=(C+C2)IAVI (5.3.2) The equivalent capacitance is then seen to be given by Coa=AV =C+C2 (5.3.3) Thus,capacitors that are connected in parallel add.The generalization to any number of capacitors is Cm=C+C2+C3++Cw=∑C (parallel) (5.3.4) i=l 9
5.3.1 Parallel Connection Suppose we have two capacitors C1 with charge Q1 and C with charge 2 Q2 that are connected in parallel, as shown in Figure 5.3.2. Figure 5.3.2 Capacitors in parallel and an equivalent capacitor. The left plates of both capacitors C1 and C2 are connected to the positive terminal of the battery and have the same electric potential as the positive terminal. Similarly, both right plates are negatively charged and have the same potential as the negative terminal. Thus, the potential difference | ∆V | is the same across each capacitor. This gives 1 2 1 2 , | | | Q Q C C V = = ∆ ∆V | (5.3.1) These two capacitors can be replaced by a single equivalent capacitor with a total charge Q Ceq supplied by the battery. However, since Q is shared by the two capacitors, we must have Q Q= +1 2 Q = C1 | | ∆V +C2 | | ∆V = ( C1 +C2 )| ∆V | (5.3.2) The equivalent capacitance is then seen to be given by eq 1 2 | | Q C C V = = +C ∆ (5.3.3) Thus, capacitors that are connected in parallel add. The generalization to any number of capacitors is eq 1 2 3 1 (parallel) N N i i C C C C C C = = + + +"+ = ∑ (5.3.4) 9