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20: Chapter Electric Potential and Capacitance 1. The concepts covered in this chapter are: 1. Electric potential 2. Electric potential energy 3. Capacitance 4. Capacitors arranged in series and parallel configuration 5. Air- and dielectric-filled capacitors 6. Potential energy stored in capacitors 2. Potential and Potential Energy. In the previous chapter we learned that charges exert a force on each other according to Coulomb's law. This means that such forces that occur among the charges can change a given configuration of charges: in the field view of electrostatic interaction the electric fields of various charges interact to make the charges move, i.e., the electric field does work on the charge configuration). The configuration can also be changed by an external agent, i.e., work can be performed by an external agent. This change in charge configuration will show up as a change in the potential energy of the system (the charge configuration). The change in the potential energy between (or of) two charges is defined as Wfield = - Wext = - U ...[1] - Uab = Ua Ub = F .dr = q E .dr a a b b ...[2] The potential difference V = Vb - Va is defined as: Vb - Va = - E .dr a b ...[3] Potential can be defined as the potential energy per unit charge ( recall a similar definition of the electric field E as the force per unit charge, qE = F) A point charge Q, creates an electric potential V(r) at a distance r from it given by: V(r) = kQ/r ( referenced to infinity, i.e., V ( ) = 0 ) ...[4] 1 The net potential at a given point due to a collection of charges is simply the algebraic sum of individual potentials V = k i =1 n Qi ri ...[5] Potential energy of two point charges is given by ...[6] U12 = kQ1Q2/r Potential energy of a collection or configuration of point charges is the algebraic sum of the potential energy of all the pairs of charges you can form in the configuration. For example, in Solved Example 1 below, Ui is the potential energy of the 3-charge configuration. Potential energy of a single charge in a collection or configuration of point charges is the algebraic sum of the potential energy of all the pairs the given charge forms in the configuration. For example, Solved Example 1 the potential energy of Q1in part [a] would be the sum U12 + U13. Points to remember: 1. Make sure you understand the difference between potential and potential energy, and between potential energy of a charge configuration and the potential energy of a charge in a charge configuration. 2. A single charge will create a potential at a point in space, but you need at least two charges for the potential energy to have any meaning. 3. Potential and potential energy are scalars. 4. Eq. [3] can also be expressed as : 2 Ex = - dV/dx ( and similar relations in the y- and z-directions) .This allows one to calculate the components of E (a vector) by knowing the potential V(x,y,z) which is a scalar. Please see sec. 20-4 of your Text for additional details. 3. Capacitance The capacitance C is defined as C = Q/V (in the text C = Q/V). C is always positive. When we talk about the charge on a capacitor we mean the magnitude of the charge on one of its plates. The capacitors can be arranged in a series or a parallel arrangement. In a series arrangement: [i] the net capacitance C is given by; 1/C = 1/C1+1/C2+1/C3+.... ...[7] [ii] the magnitude of the charge on every capacitor is the same. [iii] the potential difference across each capacitor is different and is given by Vi = Q/Ci. In a parallel arrangement: [i] the net capacitance C is given by; C = C1 + C2 + C3+.... ...[8] [ii] the potential difference V across each capacitor is the same. [iii] the charge on every capacitor is different and is given by Qi = V Ci. The capacitance of a parallel plate capacitor of plate area A and separation d is given by: C = A0/d The energy stored in a capacitor is given by U = 1/2CV2 ...[10] By using C = Q/V, eq [10] can be cast in terms of other parameters of interest. Insertion of a dielectric medium in a capacitor will change its capacitance by a factor of , where is the dielectric constant of the medium, C = C0. ...[9] 3 4. Solved Examples. Y Q1 1. Three charges are arranged at the vertices of a right-angle triangle as shown, with Q1= 4C =2Q2, and Q3 = - 3 C. [a] Determine the total potential energy of the three-charge system. Q3 53 o 5.0cm Q2 X [b] Determine the work done by an external force to move Q2 from its present position to infinity. [a] The initial potential energy, U i = 9 109 ( -12 10-12 6 10-12 8 10-12 - + ) = -2.88 J 3 10-2 5 10-2 4 10-2 [b] after Q2 has been removed to infinity, the potential energy of the remaining charges is -12 9 -12 10 ) = -3.60 J U f = 9 10 ( 3 10-2 the work done by the external force is given by W = U = Uf - Ui = - 3.60 ( - 2.88) = - 0.72J Note: in the problem above we have used W = U = Uf - Ui , where U is the change in the potential energy of the charge configuration. We could have also done the problem by looking at it from the point of view of Q2 (in a manner of speaking, charges don't really have a point of view). In this case U would be the change in the potential energy of Q2. U = Uf(Q2) - Ui(Q2) = 0 (U12 + U23) = - 9*109(- 1.2 +2.0)10-10 = - 0.72J 4 2. (Prob. 20.18 from the text) discussed in lecture This is a rather longish method but has several explanatory details. The problem can also be solved in a much more compact way. The shorter method follows the long one. (a) In an empty universe, the 20-nC charge can be placed at its location with no energy investment. At a distance of 4 cm, it creates a potential V1= 9 109 N m 2 C2 20 10-9 C ke q1 = = 4.50 kV 0.04 m r ( )( ) To place the 10-nC charge there we must put in energy U 12 = q2V1 = 10 10-9 C 4.5 103 V = 4.50 10-5 J . Next, to bring up the 20-nC charge requires energy U 23 + U13 = q3V2 + q3V1 = q3 (V2 + V1 ) 10 10-9 C 20 10-9 C = -20 10-9 C 9 109 N m 2 C2 + 0.08 m 0.04 m ( )( ) ( ) = -4.50 10-5 J - 4.50 10-5 J The total energy of the three charges is U 12 + U 23 + U 13 = -4.50 10-5 J . (b) The three fixed charges create this potential at the location where the fourth is released: 20 10-9 10 10-9 20 10-9 V = V1 + V2 + V3 = 9 109 N m 2 C2 + - C m 0.03 0.05 0.042 0.032 + ( ) V = 3.00 103 V Energy of the system of four charged objects is conserved as the fourth charge flies away: 1 1 2 2 mv + qV = mv + qV 2 i 2 f 0 + 40 10-9 C 3.00 103 V = v= 2 1.20 10-4 J 2 10 -13 ( )( ( kg )= ) 1 (2.00 10 2 -13 kg v 2 + 0 ) 3.46 10 4 m s The shorter method. [a] U = U12+ U23+ U13 = U13 = - 9*109*400*10-18/8*10-2 = - 4.5* 10-5 J (why is U12+ U23 = 0 ?) [b] (U + KE)i = (U + KE)f ( conservation of energy from last term) 5 Ui = KEf Ui = U14+ U34+ U24 = U24 = 9*109*400*10-18/3*10-2 = 12.0* 10-5 J = *2*10-13v2 (why is U14+ U34 = 0 ?) v = [12]1/2104m/s =3.46*104m/s Q1 20.0nC 4.0cm 10.0nC Q2 Q4 3.0cm 4.0cm Q3 - 20.0nC 40.0nC 3. Consider the three capacitors connected as shown. A potential difference of Vac = 120V is maintained between points a and c. 4F a b 2F 6F c [a] What is the potential difference Vbc across the 2F capacitor? [b] Determine the charge on the 4F capacitor. [c] How much energy is stored in the 6F capacitor ? [a] The 4F and the 6F capacitors can be combined into a single equivalent capacitor of capacitance, 10F. Vab = Q/10F Vbc= Q/2F or Vab/ Vbc = 2/10 = 1/5; 5Vab= Vbc Also Vab+ Vbc = 120V, solving the last two equations gives Vab=20V, and Vbc= 100V 6 [b] Q4 = Vab 4*10-6 = 8*10-5 C [c] U = (6*10-6)( Vab)2 = 1.2*10-3J 4. (prob 41 solved in lecture) (a) 1 1 1 = + Cs 15.0 3.00 Cs = 2.50 F Cp = 2.50 + 6.00 = 8.50 F 1 1 + Ceq = 8.50 F 20.0 F -1 = 5.96 F (b) Q = CV = (5.96 F )(15.0 V ) = 89.5 C on 20.0 F V = Q 89.5 C = = 4.47 V C 20.0 F 15.0 - 4.47 = 10.53 V Q = CV = (6.00 F )(10.53 V ) = 63.2 C on 6.00 F 89.5 - 63.2 = 26.3 C on 15.0 F and 3.00 F 7 Concept Questions 1. A spherical balloon contains a positively charged object at its center. (i) As the balloon is inflated to a greater volume while the charged object remains at the center, does the electric potential at the surface of the balloon (a) increase, (b) decrease, or (c) remain the same? (ii) Does the electric flux through the surface of the balloon (a) increase, (b) decrease, or (c) remain the same? Answer: (i), (b). The electric potential is inversely proportional to the radius (see Eq. 20.11). (ii), (c). Because the same number of field lines passes through a closed surface of any shape or size, the electric flux through the surface remains constant. V = ke q r [Eq. 20.11] 2. In Active Figure 20.6a, take q1 to be a negative source charge and q2 to be the test charge. (i) If q2 is initially positive and is replaced with a charge of the same magnitude but negative, does the potential at the position of q2 due to q1 (a) increase, (b) decrease, or (c) remain the same? (ii) When q2 is changed from positive to negative, does the potential energy of the two-charge system (a) increase, (b) decrease, or (c) remain the same? Active Figure 20.6 (a) If two point charges are separated by a distance r12, the potential energy of the pair of charges is given by keq1q2/r12. (b) If charge q1 is removed, a potential keq2/r12 exists at point P due to charge q2. 8 Answer: (i), (c). The potential is established only by the source charge and is independent of the test charge. (ii), (a). The potential energy of the two-charge system is initially negative due to the products of charges of opposite sign in Equation 20.13. When the sign of q2 is changed, both charges are negative and the potential energy of the system is positive. U = q1V2 = k e q1 q 2 r12 [Eq. 20.13] 3. (i) In a certain region of space, the electric potential is zero everywhere along the x axis. From this information, we can conclude that the x component of the electric field in this region is (a) zero, (b) in the +x direction, or (c) in the -x direction. (ii) In a certain region of space, the electric field is zero. From this information, we can conclude that the electric potential in this region is (a) zero, (b) constant, (c) positive, or (d) negative. Answer: (i), (a). If the potential is constant (zero in this case), its derivative along this direction is zero. (ii), (b). If the electric field is zero, there is no change in the electric potential and it must be constant. This constant value could be zero, but it does not have to be zero. 4. A capacitor stores charge Q at a potential difference V. If the voltage applied by a battery to the capacitor is doubled to 2 V, (a) the capacitance falls to half its initial value and the charge remains the same, (b) the capacitance and the charge both fall to half their initial values, (c) the capacitance and the charge both double, or (d) the capacitance remains the same and the charge doubles. Answer: (d). The capacitance is a property of the physical system and does not vary with applied voltage. According to Equation 20.19, if the voltage is doubled, the charge is doubled. C Q V [Eq. 20.19] 5. Two capacitors are identical. They can be connected in series or in parallel. (i) If you want the smallest equivalent capacitance for the combination, (a) do you connect them in series, (b) do you connect them in parallel, or (c) do the combinations have the same capacitance? (ii) Each capacitor is charged to a voltage of 10 V. If you want the largest combined potential difference across the combination, (a) do you connect them in series, (b) do you connect them in parallel, or (c) do the combinations have the same potential difference? Answer: (i), (a). When connecting capacitors in series, the inverse of the capacitances add, resulting in a smaller overall equivalent capacitance. (ii), (a). When capacitors are connected in series, the voltages add, for a total of 20 V in this case. If they are combined in parallel, the voltage across the combination is still 10 V. 9 6. You have three capacitors and a battery. In which of the following combinations of the three capacitors will the maximum possible energy be stored when the combination is attached to the battery? (a) When in series the maximum amount is stored. (b) When parallel the maximum amount is stored. (c) Both combinations will store the same amount of energy. Answer: (b). For a given voltage, the energy stored in a capacitor is proportional to C: U = C(V)2/2. Therefore, you want to maximize the equivalent capacitance. You do so by connecting the three capacitors in parallel so that the capacitances add. 10

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