Download 2102 Fall 97 Test 1

2102 Fall 97 Test 1
Fall 97 Test 1, P. 2
Fall 97 Test 1, P. 3
Fall 97 Test 1, P. 4
Fall 97 Test 1, P. 5
5. (10 points) A spherical rubber balloon has a charge uniformly distributed
over is surface. The balloon is slowly inflated and remains spherical during
inflation.
a) At a point well away from the balloon, does the electric field increase,
decrease, or stay the same as the balloon inflates. Explain why. Use
only words in complete sentences to explain your answer; no formulas,
symbols or diagrams.
At a point well away from a spherical symmetric charge, the
electric field is independent of the size of the sphere and
therefore will stay the same as the balloon inflates. A
point charge, a spherical shell of charge, a charge distributed
uniformly throughout a spherical volume all produce the
same field if the total charge is the same. The easiest way
to see this is with Gauss’ law. A spherical gaussian surface
through a point away from the surface will have the same
flux going through it no matter what the radius of the
balloon on the inside.
b) At the outer surface of the balloon, does the electric field increase,
decrease, or stay the same as the balloon inflates. Explain why. Use
only words in complete sentences to explain your answer; no formulas,
symbols or diagrams.
Since the distance from the center to the outer surface of
the balloon increases as the balloon inflates, the electric
field will decrease as the balloon inflates. Again using
Gauss’ law with a spherical gaussian surface incrementally
outside the surface, the flux which is field times area must
stay the same since the charge inside is constant. Since the
area increases, the electric field must decrease in order for
flux to be constant.
c) At a point inside the balloon, does the electric field increase, decrease, or
stay the same as the balloon inflates. Explain why. Use only words in
complete sentences to explain your answer; no formulas, symbols or
diagrams.
By Gauss’ law, the field everywhere inside a spherical shell of
charge is zero. Take a spherical gaussian surface with a
radius less than that of the balloon. There is no charge
inside, therefore the flux is zero meaning there can be no
electric field. The field inside stays the same (equal t o
zero) as the balloon inflates.
Fall 97 Test 1, P. 6
6. (10 points) Describe how you would use Gauss’ law to calculate the electric
field at point P between the two charged infinitely long cylinders shown with
linear charge densities of λ1 and λ 2 . Describe:
• the gaussian surface you would use (size, shape, orientation).
• how you would calculate the flux through your gaussian surface.
• how you would determine the charge inside your gaussian surface.
Use only words and proper English in your answer; no formulas, symbols or
diagrams.
In this case the
gaussian surface will
be a cylinder
through the point,
concentric with the
charged cylinders. It
will be an arbitrary
length. Since the
charged cylinders
are infinitely long,
the electric field
between the
cylinders is radial
and perpendicular to
the axis. This means
that the electric
field is perpendicular
to the curved
surface of the
cylinder.
P
λ1
λ2
The flux through this part of the gaussian surface is simply
the electric field times the area of the surface or the
circumference times the length. The flux through either end
cap of the gaussian cylinder is zero since the electric field is
parallel to the surface. On the other side of Gauss’ law we
have to calculate the charge inside the cylinder. This is
simply the linear charge density on the inner cylinder times
the arbitrary length. Equating these two, the arbitrary
length cancels out leaving electric field equal to the linear
charge density divided by the permittivity of free space
times the circumference of the gaussian cylinder.
Fall 97 Test 1, P. 7
7. You are given a parallel plate capacitor with square plates of area A, and
separation d, in a vacuum. Each plate has a charge of magnitude q. There is
no battery nor any other source of charge connected to the plates. If the
plate area is doubled, explain what happens to:
• the surface charge density,
• the electric field between the plates,
• the capacitance.
Explain the physics behind your answers. Use only words in complete
sentences. Do not use formulas, symbols, or diagrams in your answer.
The surface charge density, being the charge divided by the
area, simply decreases by a factor of 2 if the area is
doubled. The charge that is present must now spread itself
out over an area that is twice as large. The electric field
between the plates, therefore, will decrease by a factor of
two since it is directly proportional to the surface charge
density. The capacitance will now be a factor of two larger
since the charge on the plates is the same but the potential
is a factor of two lower due to the lower electric field. Since
the capacitance is the ratio of the charge to the potential
applied to the plates, and the potential is a factor of two
lower, the capacitance must be a factor of two larger.