What is the flux through the cylinder shown at left, inside the capacitor?

1

• What is the flux through the cylinder shown at
  left, inside the capacitor?
• E A, because the field strength is constant
  inside the capacitor
• 0, because there is no charge inside the cylinder
• ½ E A, because the field lines coming in on one
  side cancel some of those coming out on the other
• 2 E A, because there are two cylinder ends
  between the plates there is twice the flux
  through the cylinders surface.

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2
Correct Answer B Gauss Law tells us that the
total flux through a closed surface is equal to
the charge inside the surface (divided by the
electrical permitivity). Since the closed
surface depicted is entirely within the
insulating region between the plates, there is no
charge inside it. Therefore the flux through it
must be zero. How can we understand this in terms
of flux lines? There clearly are flux lines
entering through the left hand end of the
cylinder. But everyone of those flux lines exits
through the right hand end, and each line coming
out cancels each line coming in.
3

• What is the total flux passing through the
  cylinder shown? s is the surface charge density
  on each plate
• 2 s A, because that is the amount of charge
  inside the cylinder
• s A, because only one of the plates contributes
  to the flux through the cylinder
• ½ s A, because the negative charges on one plate
  cancel some of the charges on the other plate
• 0, because there are no flux lines passing
  through the cylinder

- s
s
4
Correct Answer D If you look carefully at the
diagram, you see there can be no flux lines
passing through the closed surface of the
cylinder. The curved surface of the cylinder is
parallel to the flux lines, so none of the lines
pass through this surface. In this cylinder the
end surfaces are outside the capacitor, and we
know that the flux lines, which cannot pass
through the conducting plates, are not present
outside the capacitor. So if the total flux
through the closed surface is zero, Gauss Law
says the total charge inside it must be zero.
But we do know there is charge on the capacitor
plates. The key here is that equally sized parts
of both plates are inside the surface, so there
must be an equal charge density on the two
plates. Since one plate is positive and one plate
is negative, the equal charge amounts cancel each
other out.
5

• Suppose we shrink the length of the negative
  plate to a half that of the positive plate. Where
  is the field strength greatest inside the
  capacitor?
• Close to the positive plate (longer plate)
• In the middle
• Close to the negative plate (shorter right)
• It is still the same everywhere inside the
  capacitor

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6
Correct Answer C A careful look at the diagram
shows clearly that the flux lines are closer
together closer to the smaller negative plate.
Closer together flux lines correspond to stronger
fields. The fact that the flux lines from the
positive plate all end on the negative plate
tells us that the two plates still have equal
total charges Q. But since the area of the
negative plate is smaller, the charge density on
that plate is larger. Since electric field inside
the capacitor depends directly on the charge
density on the plates (see last pages of chapter
19) it follows that in the case of unequally
sized plates the plate with greater charge
density should have a greater electric field
strength close to it.
7

• In the case of the unequal plates, if we again
  draw a closed cylinder whose ends reach beyond
  the plates, what can we say about the total
  charge inside the cylinder?
• It is zero
• It is positive
• It is negative
• It is infinite

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8
Correct Answer C In the previous question we
stated that the charge density on the smaller
negative plate is greater than that on the larger
positive plate. Since the cylinder has equal
areas of the two plates within it, it follows
there must be more negative charge inside the
cylinder than there is positive charge. Therefore
the total charge inside the cylinder is
negative. What about the flux lines? There is
still no flux through the ends of the cylinder,
but the flux lines are no longer parallel to the
curved surface of the cylinder. Since the flux
lines all come diagonally into the cylinder from
the outside it follows ther eis a net negative
flux into the cylinder.
9

• What is the field strength in the middle of the
  capacitor, equidistant from the two plates, for
  the flux lines drawn? Let E be the field
  strength close to the positive plate (the longer
  plate), and E- be the field strength close to the
  negative plate (the shorter plate).
• 0
• E-
• ½ (E- E)
• E
• E- E

-

10
Correct Answer C The flux lines are all drawn
as straight in the illustration (this is not
quite realistic, but is good enough for our
purposes). Since they are straight, the change in
field strength must be linear, so it follows that
the field strength in the center must be the
average of the field strength at the plates. What
might be more realistic? We know field lines
repel each other. When they are close together
they repel more strongly, so the field lines can
draw closer together more quickly near the
positive plate where the field is weaker. This
suggests the point where the field strength is
halfway between the values at the plates might be
a little closer to the positive (larger) plate.
11

• If a charge q is placed inside the capacitor,
  what is the amount of the force, F, propelling it
  towards the negative plate? (A is the area of the
  plate, r is the distance between the charge and
  the positive plate, s is the surface charge
  density on each plate, e is the electrical
  permittivity of free space).
• F q E q s / e
• F k q Q /r2 k q s A/r2
• F k q Q /r2 k q Q /r2 2 k q s A/r2
• F k q Q /r2 k q Q /r2 0

F
q
- s
s
12
Correct Answer A We know that a charge q placed
in an electric field E experiences a force F q
E. Inside the plates of the capacitor with charge
density s, the field strength is E s / e
13

• Suppose we take a rectangular loop of wire and
  place it inside the capacitor as shown. Given
  that the field strength is constant everywhere in
  this capacitor, is there a way to move the wire
  so that the amount of flux passing through it
  changes?
• No, because the field is constant inside the
  capacitor.
• Yes, we just have to rotate the loop

14
Correct Answer B Its true that if we just move
the loop up or down and back or forth that the
flux through it wont change. For each line that
passes inside the loop, another line leaves,
because the flux is constant inside the
capacitor. But if we rotate the loop, matters are
different. When the loop is parallel to the flux
lines, they dont actually pass through it, and
the flux through it is zero. When it is
perpendicular to the flux lines, they penetrate
it maximally, and the flux through it is equal to
E A where A is the area of the loop. So by
rotating a loop we can change the flux through
it, even when the field is constant. This will be
important later on when we look at magnetic
fields.