üA If a spherical raindrop of radius 0 650 mm by xyh18958

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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

								
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