Physics 304
Problem Set 2
Due: Monday, Sept. 12
(1) Griffiths problem 1.56 (page 55).
(2) A hollow spherical shell carries charge density ! = Ar between r = R and r = 2R,
with A a constant. For r < R and r > 2R there is no charge.
(a) What is the total charge?
(b) Using the integral form of Gauss’ law, find the electric field in the regions r < R, R <
r < 2R and r > 2R.
(c) From your result in (b), integrate to find the electric potential in all three regions.
Sketch E and V.
(d) Using the differential form of Gauss’ law, verify that your electric field is consistent
with the given charge density within the shell for R < r < 2R, as well as outside the shell.
(3) An infinite slab of charge, with thickness 2d, carries a charge density ! = " z 4 in the
region between z = –d and d, and zero charge elsewhere. The parameter α is a constant.
(a) By symmetry, E will be zero at z = 0, and the x and y components will vanish. Use
!Ez
the differential Gauss’ law to find
!z both inside and outside the slab. Integrate to
find the electric field in all 3 regions: z < –d, –d < z < d, and z > d.
(b) The electric field on the two sides of the slab must also be consistent with the integral
form of Gauss’ law. Demonstrate this by using a pillbox that completely surrounds the
slab.
(4)
to exist in a region surrounding the origin:
! Consider two fields proposed
!
E1 = C [ zx̂ + yŷ + xẑ ] and E2 = C [ zx̂ + yŷ ! xẑ ] . Which field could exist in electrostatics?
For the field that could exist, find the charge density, and find the potential relative to the
origin. Note that to find the potential you can start at the origin and choose a general path
to a position (x, y, z). (The field will become infinite at very large distances, which
indicates that this field may exist only in a finite region, and this is why we choose the
origin as the reference point.)
(5) Find the total energy required to assemble the charge distribution described in
problem 2 above.
(6) Consider a long hollow tube consisting of a very thin insulating layer at radius R,
with charge density ! = A cos" , where A is a constant. The length can be considered to
be infinite. As can be shown by methods in chapter 3, the electric field inside will be
uniform: Ein = ! x̂ , while the field outside will correspond to an electric potential of the
form, Vout = (! cos" ) s .
(a) Find the cylindrical components of the electric field outside the tube starting with the
potential defined above, and then use the pillbox method at the edge of the tube to
establish a relationship between α and β.
(b) Use a direct integration method similar to equation 2.29 in the text to find the electric
potential at the origin, relative to infinity, for this charge density.
(c) The electric potential can also be found by integrating the electric field, as in
equation 2.21. Using this method and the inside and outside fields defined above, solve
for !V from the origin to infinity. Since your result should be equivalent to that of part
(b), you can establish a second relationship between α and β in this way. Do this, and
finally solve for the constants α and β by combining your results.