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Problem 23.80 A. If a spherical raindrop of radius 0.650 mm carries a charge of -1.70pC uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.)? Since the potential is zero at infinite distance from the raindrop we only need to consider the potential of the drop itself. Since it is spherically symmetric we know that the potential will be of the same form as that of a point charge located at the origin. This point charge will be equal to the total charge of the raindrop. Answer in V. q V 4 Π Εo r 12 1.7 10 12 4 4Π 8.85 10 6.5 10 23.517 B.Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?? Take V and R to be the volume and radius of the two original raindrops and Vnew and Rnew to be the volume of the new raindrop created from the collision of the two original ones. We can then set Vnew = 2V, substitute in the following expression for volume in terms of radius and solve for Rnew in terms of R Answer in m. 4 Π R3 4 Π Rnew3 V , Vnew 3 3 3 2V Vnew 2 R3 Rnew3 2 R Rnew 3 4 2 6.5 10 0.000818949 C.What is the potential at its surface, if its charge is uniformly distributed over its volume?? We can use the same equation and setup as in part A but we must use Rnew for r and double q to reflect the increase in charge from the combined raindrops. Answer in V. 2q V 3 4 Π Εo 2 r 12 1.7 10 2 3 12 4 4Π 8.85 10 2 6.5 10 37.331 Potential of a Charged Disk 2 solutions_2_v2.nb Potential of a Charged Disk A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in the xy 1 plane. Throughout this problem, you may use the variable k in place of 4 Π Ε0 A. What is the electric potential V(z) on the z axis as a function of z , for z > 0 ? Since we can add up the individual contributions to the potential from each piece of the charged disk to find the total potential at a given location we will find V(z) by integrating over the entire disk. This is most easily accomplished in polar (cylindrical) coordinates. For and arbitrary point{0, 0, z} on the z-axis and an arbitrary point {r, Θ, 0} in the disk we can write the distance between them as z2 r2 . This yields the contribution to the potential, dV from one point in the disk. Σ dV k z2 r2 Q where Σ is the surface charge density of the disk given . Integrating in polar coordinates and making sure to use the approriate Π a2 area element dA = r dr dΘ. Note that the symmetry of the problem about the z-axis demands an answer that is independent of Θ, which is confirmed by the solution of the integral. a 2Π Σ dV r Θ r 0 0 k z2 r2 2kQ V z z2 a2 z a2 B.What is the magntude of the electric field E on the z axis as a function of z , for z > 0 ? We know from symmetry that E at a point on the z-axis will point along the z-axis and thus have no components in either the r or V Θ (x or y) directions. Therefore we can use E = - z to get Q z E z 1 2 Π Ε0 a2 z2 a2 Exercise 23.4 A. How much work would it take to push two protons very slowly from a separation of 10 15 2*10 m (a typical atomic distance) to 3*10 m (a typical nuclear distance)? Answer in J To find the Work need to move the proton from point A to point B we can find the difference in the potential of the two configurations qproton qproton 4 Π Εo A 4 Π Εo B solutions_2_v2.nb 3 19 2 1.6 10 1 1 12 10 15 4Π 8.85 10 2 10 3 10 14 7.67289 10 B.If the protons are both released from rest at the closer distance in part A, how fast are they moving when they reach their original separation?Answer in m/s Use conservation of energy. From part A. we know the change in potential energy between the closer distance and the original distance. If this potential is converted into the kinetic energy of both particles we know that 1 U U 2 mproton v2 v 2 mproton 14 7.68 10 27 1.67 10 6.78145 106 Exercise 23.49 A metal sphere with radius ra is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb . There is charge +q on the inner sphere and charge -q on the outer spherical shell. Take V to be zero when r is infinite. A. Calculate the potential V(r) for r < ra . (Hint: the net potential is the sum of the potentials due to the individual spheres.)? 1 1 V r kq ra rb B.Calculate the potential V(r) for ra r rb ? 1 1 V r kq r rb C.Calculate the potential V(r) for r rb ? Net charge is 0 so V r 0 4 solutions_2_v2.nb D. Find the potential of the inner sphere with respect to the outer? 1 1 V r kq ra rb V E. Use the equation E r and the result from part (b) to find the electric field at any point between the spheres? Take the derivative of the answer from part B with respect to r. D kq 1 r 1 rb , r kq r2 V F. Use the equation E r and the result from part (c) to find the electric field at any point r > rb ? Taking the derivative of the answer from part c with respect to r we get 0. G. Suppose the charge on the outer sphere is not -q but a negative charge of -Q different magnitude, say . Find the potential of the inner sphere with respect to the outer. 1 1 V r kq ra rb H. Suppose the charge on the outer sphere is not -q but a negative charge of -Q different magnitude, say . Find the electric field at any point between the spheres (ra r rb ). D kq 1 r 1 rb , r kq r2 I. Suppose the charge on the outer sphere is not -q but a negative charge of -Q different magnitude, say . Find the electric field at any point between the spheres (rb r). Add the contributions to the E field from each sphere. kq kQ kq Q 1 r2 r2 r2 q Electric Fields and Equipotential Surfaces The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1-V increments. A. What is the work Wab done by the electric force to move a 1-C charge from A to B? Answer in J solutions_2_v2.nb 5 A. What is the work Wab done by the electric force to move a 1-C charge from A to B? Answer in J Since points A and B are one the same equipotential surface they are by definition at the same potential and hence there is not potential difference between them and no work done moving a charge between them. B. What is the work Wad done by the electric force to move a 1- charge from A to D? Answer in J Va Vd 1 C. The magnitude of the electric field at point C is a. greater than the magnitude of the electric field at point B. Since equal amounts of work are done to move a charge from one equipotential surface to an adjacent one it follows then that if the equipotentials are closer together, the electric force does the same amount of work in a smaller displacement than if the equipotentials were farther apart. Therefore, the electric force, as well as the corresponding electric field, has a larger magnitude. Another way to think about this is that equipotentials are like contours on an elevation plot. Therefore where the contours are closer together the elevation (analogous to V) is changing over a smaller distance (the distance between contours) and the V topography is thus steeper there. Since E r we know that the electric field strength is analogous to the slope ('steepeness') of the potential at a given point. Capacitance and Electric Field of a Spherical Capacitor A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has a radius of ra = 12.3 cm , and the outer sphere has a radius of rb = 14.8 cm. A potential difference of 120V is applied to the capacitor. A. What is the capacitance of the capacitor? Answer in F. Put +Q on the inner spherical shell and -Q on the outer. Then as we know from 23.49 the field in the region between the shells will point in the radial direction and be given by Q E 4 Π Ε0 r2 rb and the corresponding potential, also from 23.49 (or integrate V=- ra E l ) will be given by Q 1 1 V 4 Π Ε0 ra rb Q Capacitance, C, is defined as V so we can write 6 solutions_2_v2.nb ra rb C 4 Π Ε0 rb ra 12 .123 .148 4 Π 8.85 10 .148 .123 11 8.09804 10 B.What is the magnitude E1 of the electric field E at radius 12.7 cm, just outside the inner sphere? Answer in V/m First find Q in units of C from the equation for V in part A. 4 Π Ε0 V Q 1 1 ra rb 12 4 Π 8.85 10 120 1 1 .148 .123 9 9.71765 10 Now find E from the equation in part A. evaluated at a radius of 12.7 cm. 9.72 10 ^ 9 12 4 Π 8.85 10 .127 ^ 2 5418.83 C. What is the magnitude of E at radius 14.7 cm , just inside the outer sphere? Answer in V m 9.72 10 ^ 9 12 4 Π 8.85 10 .147 ^ 2 4044.63 Equivalent Capacitance Consider the combination of capacitors shown in the diagram, where C1 = 3.00ΜF , C2 = 11.0ΜF , C3 = 3.00ΜF , and C4 = 5.00 ΜF . A. Find the equivalent capacitance Ca of the network of capacitors? Answer in ΜF Recalling that the rules for finding equivalent capacitances are like those for resistors except we switch the cases for parallel and series. 1 1 1 Ca C1 C2 C3 C4 solutions_2_v2.nb 7 1 1 1 N 3 11 3 5 2.59091 B.Two capacitors of capacitance C5 = 6.00 and C6 = 3.00 are added to the network, as shown in the diagram. (Part B figure) Find the equivalent capacitance C5 of the new network of capacitors? Answer in ΜF 1 1 1 1 1 1 where C26 CB C1 C3 C4 C5 C26 C2 C6 1 1 1 In[1]:= N 3 1 1 1 3 5 6 11 3 Out[1]= 2.53506 Exercise 24.33 For the capacitor network shown in the Figure , the potential difference across ab is 220 V. A. Find the total charge stored in this network? Answer in ΜC Q VC 9 220 35 75 10 In[9]:= Q ; 6 10 N Q Out[10]= 24.2 B. Find the charge on each capacitor? Answer in ΜC Q VC 9 220 35 10 In[11]:= Q ; 6 10 N Q Out[12]= 7.7 9 220 75 10 In[13]:= Q ; 6 10 N Q Out[14]= 16.5 8 solutions_2_v2.nb C. Find the total energy stored in the network.? Answer in mJ 1 W C V2 2 1 9 In[21]:= W 110 10 2202 1000; 2 N W Out[22]= 2.662 D. Find the energy stored in capacitor? Answer in mJ 1 W C V2 2 1 9 In[23]:= W 75 10 2202 1000; 2 N W Out[24]= 1.815 1 9 In[25]:= W 35 10 2202 1000; 2 N W Out[26]= 0.847 E. Find the potential difference across each capacitor? Answer in V V 220 for both capacitors since they are in parallel between a and b A Simple Network of Capacitors In the figure are shown three capacitors with capacitances C1 = 6ΜF, C2 = 3ΜF, C3 = 5ΜF. The capacitor network is connected to an applied potential Vab . After the charges on the capacitors have reached their final values, the charge Q2 on the second capacitor is 40.0 ΜC. A. What is the charge Q1 on capacitor C1? Answer in ΜC Q C1 and C2 are in parallel so we know that they have the same voltage drop across them (V1 = V2). Using V= C for each capaci- tor independently we can write Q1 Q2 C1 C2 and solve for Q1 solutions_2_v2.nb 9 6 In[29]:= Q1 40 3 Out[29]= 80 B.What is the charge on capacitor C3? Answer in ΜC Since all the capacitors are initially charge neutral and charge is conserved, we know that the charged capacitors in series must have a net charge of 0. The charge on the first two capacitors (in parallel) sums to 120ΜC therefore the charge on C3 must balance this and be equal to 120ΜC. Since the positive plate of one capacitor is connected to the negative plate of the other the charges are opposite in sign in the circuit. C. What is the applied voltage Vab ? Answer in V Q First we need to find the equivalent capacitance of the whole circuit. Then use V = C to find Vab 1 1 1 Ctot C3 C1 C2 1 1 1 In[32]:= Ctot N 5 6 3 Out[32]= 3.21429 120 In[33]:= V Ctot Out[33]= 37.3333 Problem 24.61 Three capacitors having capacitances of 8.5ΜF, 8.4ΜF, and 4.3ΜF are connected in series across a 40 V potential difference.? A. What is the charge on the 4.3 capacitor? Answer in C Since the capacitors are in series we know that Q should be the same for all of them: Using the equivalent capacitance of the circuit and Q= V C V Q1 1 1 1 C1 C2 C3 40 6 In[37]:= 10 1 1 1 8.5 8.4 4.3 Out[37]= 0.0000852419 10 solutions_2_v2.nb B.What is the total energy stored in all three capacitors? Answer in J 1 1 V2 2 2 U CV 2 1 1 1 C1 C2 C3 .5 402 10 6 In[39]:= 1 1 1 8.5 8.4 4.3 Out[39]= 0.00170484 C.The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? Answer in V Using V =Q C and the Q found in part A. for the charge on each of the 3 capacitors and the equivalent capacitance of three in parallel C = C1 + C2 + C3. 3V V C1 C2 C3 1 1 1 C1 C2 C3 5 3 8.5 10 In[42]:= V 6 8.5 8.4 4.3 10 Out[42]= 12.0283 D.What is the total energy now stored in the capacitors? Answer in J Using the results from the previous part of the problem for V and Ctot and the equation for the energy of the capacitor. 1 U C V2 2 5 2 3 8.5 10 6 In[44]:= 1 2 8.5 8.4 4.3 10 6 8.5 8.4 4.3 10 Out[44]= 0.00153361