ANSWER TO CHAPTER 24 HOMEWORK

1. We assume that the charge transferred is small compared to the initial charge on the plates so the potential difference between the plates is constant. The energy required to move the  charge is

            W = qV.

         Thus the charge on each plate is

            Q = CV = C(W/q) = (16 x10-6F)(25J)/(0.20 x 10-3C) =  2.0C.

         Because this is much greater than the charge moved, our assumption is justified.

2.  The uniform electric between the plates is related to the potential difference across the plates:

            E = V/d.

         For a parallel-plate capacitor, we have

            Q = CV = (eoA/d)(Ed) = eoAE;

            4.2 x 10-6C = (8.85 x10-12C2/N.m2)A(2.0 x 103V/mm)(103mm/m).

         which gives  A = 0.24m2.

3.  (a) From the circuit, we see that C2 and C3 are in series and find their equivalent capacitance from

            1/C4 = (1/C2) + (1/C3), which gives C4 = C2C3/(C2 + C3).

         From the new circuit, we see that C1 and C4 are in parallel, with an equivalent capacitance

             Ceq = C1 + C4 = C1 + [C2C3/(C2 + C3)]

                    =  (C1C2 + C1C3 + C2C3)/(C2 + C3).

        (b) Because V is across C1, we have

              Q1 = C1V = (14.0mF)(25.0V) =  350mC.

             Because C2 and C3 are in series, the charge on each is the charge on their equivalent capacitance:

              Q2 = Q3 = C4V = [(C2C3/(C2 + C3)]V

                    = [(14.0mF)(7.00mF)/(14.0mF + 7.00mF)]/(25.0V) =  117mC.

            

  
4.  Because C1 and C2 are in series, we have

                Q1 = Q2 = 12.4mC.

         Thus we have

                V1 = Vcd = Q 1/C1 = 12.4mC/16.0mF = 0.775V;

                V2 = Vdb = Q2/C2 = 12.4mC/16.0mF = 0.775V.

         From the diagram we see that

                V3 = Vcd + Vdb  = 0.775V + 0.775V = 1.55V, so

                Q3 = C3V3 = (1.60mF)(1.55V) = 24.8mC.

         For Q4 we have

                Q4 = Q1 + Q3 =12.4mC + 24.8mC = 37.2mC, so

                V4 = Vac = Q4/C4 = 37.2mC/36.0mF = 1.03V.

         From the diagram we see that

                Vab = Vac + Vcb = 1.03V + 1.55V = 2.58V.

         Thus we have 

                Q1 = Q2 = 12.4mC, Q3 = 24.8mC, Q4 = 37.2mC;

                V1 = V2 = 0.775V, V3 =1.55V, V4 =1.03V, Vab  = 2.58V.

  

5.  (a) When the capacitors are connected in parallel, we find the equivalent capacitance from

                     Cparallel = C1 + C2 = 0.15mF + 0.20mF = 0.35mF.

               The stored energy is

                     Uparallel = 1/2CparallelV2 = 1/2(0.35 x10-6F)(12V)2 = 2.5 x 10-5J.

           (b) When the capacitors are connected in series, we find the equivalent capacitance from

                     1/Cseries = (1/C1) + (1/C2) = [1/(0.15mF)] +1/(0.20mF)], which gives Cseries = 0.0857mF.

                 The stored enegy is

                     Useries = 1/2Cseries V2 = 1/2(0.0857 x 10-6F)(12V)2 = 6.2 x 10-6J.

            (c) We find the charges from

                     Q = CeqV;

                     Qparallel = CparallelV = (0.35mF)(12V) = 4.2mC.

                     Qseries = CseriesV = (0.0857mF)(12V) -= 1.0mC.

6.  the potential difference must be the same on each half of the capacitor, so we can treat the system as two capacitor is parallel:

                    C = C1 + C2 = [K1eo(1/2A)/d] + [K2eo(1/2A)/d]

                        = (e01/2A/d)(K1 + K2) = 1/2(K1 + K2)(eoA/d)

                        =  eoA(K1 + K2)/2d.

  
 
7*. For a small angle q,  we have tanq  » q. If we consider a differential element a distance y from the small end, the capacitance of the element is

                     dC = eodA/(d + y tanq) » eoldy(d + yq),

where l is the length of a plate. the infinite number of elements are in parallel, so we find the total capacitance by integrating:

                     C = eoò0ldy/(d + yq)

                        = eol/q ln[(d + yq)/d]

                        = eol/q ln(1 + lq/d).

We use the expansion ln(1 + x) » x -1/2x2, for small x:

                   C = (eol/q)[(lq/d) -1/2(lq/d)2]

                       = (eol2/q)[1 - 1/2(lq/d)] = (eoA/d)[1 - 1/2(qÖA/d)]].

 
 
8*. (a) We can treat the system as a capacitor of length l - x in parallel with a capacitor of length x:

                   C = Cair + Cdielectric

                       = [eol(l - x)/d] + [Keolx/d]

                       = (eol2/d)[1 - (x/l) + K(x/l)]

                       = (eol2/d)[1 + (K - 1)(x/l)].

(b) the energy stored is

                   U = 1/2CV02 = (eol2V02/2d)[1 + (K - 1)(x/l)].

 (c) When the slab moves a small distance dx, the capacitance change is 

                   dC = (eol2/d)(K -1)dx/l = eol(K - 1)dx/d.

Because the voltage is constant, this increase in capacitance means a decrease in the charge on the capacitor plates, which means some charge is returned to the voltage source. The magnitude of the change in charge is

                  dQ = V0dC = eolV0(K - 1)dx/d.

We see from part (b) that there is an increase in the energy stored in the capacitor:

                  dUcapacitor = eolV02(K - 1)dx/2d.

For the entire system we must include the decrease in energy in the voltage source:
                  dUsource = -V0dQ = -eolV0(K - 1)dx/d.

We find the external force required to produce the total energy change from

                  dW = dUcapacitor + dUsource 

                  Fdx = eolV02(K - 1)dx/2d - eolV02(K - 1)dx/d

                          = -eolV02(K - 1)dx/2d.

Thus the external force is

                 Fexternal = -eolV02(K - 1)/2d, that is, to the right to oppose the attraction between the charges on the plates and the induced charges on the dielectric, and thus keep the slab from accelerating.

Thus the force of attraction on the slab is

                 Fslab  = +eolV02(K - 1)/2d, to the left, drawing the slab between the plates.

 
9*. We put a linear charge density of - l on the innermost cylinder and + l on the outermost cylinder. Because the two cylinders between these are connected, they will acquire equal and opposite charge densities so that they are at the same potential and have no electric field in the region between them. A charged cylinder has no electric field within it and an electric field outside it that is the same as that of a line charge:

                  E = l/2peor.

We find the potential difference between c and d by integration:

                 Vc - Vd = - òRdRcE·dl = - òRdRc(l/2peor)dr

                              = +(l/2peo)ln(Rc/Rd).

The electric field between a and b will be due to the net line charge. We integrate to find the potential difference between a and b:

                 Va - Vb = - òRbRaE·dl = - òRbRa(l/2peor)dr

                              = +(l/2peo)ln(Ra/Rb).

Because Vb = Vc, we can use these results to get the potential difference across the capacitor:

                 Va - Vd = (Va - Vb) + (Vc - Vd)

                              = +[(l/2peo)ln(Ra/Rb)] + [+(l/2peo)ln(Rc/Rd)]

                              = (l/2peo)ln(RaRc/RbRd).

Thus the capacitance per unit length is 

                 C/L = Q/LVad = 2peo/ln(RaRc/RbRd).

Note that this is also the result if the arrangement is treated as two cylindrical capacitors in series.

 

 

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