1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A rod and two balloons

  1. Feb 21, 2004 #1
    I'm currentlt taking physics 4b (electricity and magnetism) and I'm having problems deciphering a homework problem.

    I'm just mystified at what the question is asking, could someone restate the question in simpler terms? I have copied the question verbatim, I will attempt to supply additional information.
  2. jcsd
  3. Feb 21, 2004 #2
    Were you given a diagram along with that assignment? They seem to be asking you to calculate the direction of the electric field, but at which point? The field direction will vary depending on where you are in relation to those charged objects.

    And from that description, you can't even tell how the objects are oriented. It could look like this:
    o o

    or this:
    o . o
    (here the line is perpendicular to your screen)

    If you know how they are arranged, and you have a point P that you're dealing with, then calculate the field (expressed as a vector) at P from each of the three objects, & sum the three vectors.

    If P is very near the center of the rod, you are probably safe to assume that the rod is infinite in length -- I imagine that's the approximation they alluded to.
  4. Feb 21, 2004 #3
    I cannot post the diagram for some reason, but I will do my best to explain. The rod is lying horizontally, the balloons are lying 3cm away from the center of the rod (from the edge of the rod to the centers of the balloons) and are 2cm apart (from edge to edge).

    Part 'A' says:

    "Find the magnitude of the electric field at the location marked 'X', .6 cm to the right of the surface of the left balloon."

    The mark in the figure is shown as the center of the left most balloon (.6cm from its surface, radius = 1.2). The mark lies in the plane of the center of the other balloon and is perpendicular to the rod.

    Sorry for not including that, the question means so little to me, I was confused as to what to include.
  5. Feb 21, 2004 #4
    I was able to copy the image from the homework problem. It has to be in zip format, I don't know why the forum code won't allow me to attach a 630x432 image when its well within the maximum attachment size. But whatever there it is.

    Attached Files:

  6. Feb 21, 2004 #5
    That helps.

    X is inside the left balloon, so at X, the field due to the left balloon is 0 (review Gauss's law if you're unclear about this).

    So you can ignore that balloon.

    Compute the field at that point arising from the right balloon (which will be the same as if the right balloon were a point charge of the same magnitude located at its own center). The balloon has a negative charge. What is the direction of that field?

    Also compute the field at X arising from the glass rod. That has a positive charge. What is the direction of its field?

    Add the two fields (the resultant of the two vectors).

    Once you have an expression for the resultant vector, you can determine the angle it makes with the x-axis.
  7. Feb 21, 2004 #6
    Thank you that's much clearer.

    I am unclear on Gauss's law apparently:

    Why is this? My teacher is unclear at best, so you'll have to bear with me as I stumble through this. Thanks for your help.
  8. Feb 21, 2004 #7
    Well, I'm afraid I can't give you a rigorous lesson in Gauss's law, but I'll try to help a bit.

    It tells us that if you imagine any closed surface (i.e. a hollow sphere, a hollow cube, etc) the net electric flux ΦE through the entire surface (i.e. the number of lines of force entering the surface minus the number of lines of force exiting the surface is proportional to the total charge enclosed by the surface:
    [tex] \Phi_E = \frac{q}{\epsilon_0} [/tex]
    where ε0 is a constant called the permittivity of free space.

    In situations where there is sufficient SYMMETRY so that you can argue that the flux is the same through every point on a particular surface (often an imaginary surface), that provides a convenient way to determine the strength of the field at that surface. In your situation, if you imagine a sphere inside your balloon, concentric with it but with a smaller radius so that point X lies on the surface of the imaginary sphere (the "Gaussian surface"), it is easy to see that there is NO charge enclosed by this sphere, and (maybe not quite as easy to see) because of the symmetry of two concentric spheres, any field caused on the inner sphere by the charge on the outer sphere must be the same everyplace on the surface of the inner sphere. So those two facts -- the net flux through the entire surface of the inner (imaginary) sphere is zero, and the flux through every point on that imaginary sphere is the same as every other point on that imaginary sphere, means that the net flux (and hence the field) must be zero everyplace on the inner sphere.

    I hope that helps (and doesn't confuse you more).

    Note that this only works where there is sufficient symmetry, so this argument ONLY applies as to the field caused by the left balloon.
  9. Feb 21, 2004 #8
    Here's another way to look at it.

    First think of the center point of the balloon (sphere). A charge placed at that point will feel an electric field emanating from every point on the surface, but since every point is the same distance away, and evenly distributed in all directions, the NET field at the center is clearly 0, right?

    Now think of another point, inside but close to the inner surface of the balloon. A charge placed there, will feel the electric field from points nearby on the surface strongly, since it is close to those points, and weakly from points located farther away on the surface. But there are many more points farther away, enough so that their combined influence, even though weaker individually, exactly counterbalances the smaller number of nearby points. If you did an integral over the entire surface computing the total force exerted by every little charge "dq" on a charge placed at your point X, the result would be 0.
  10. Feb 22, 2004 #9
    So due the symmetry of the sphere any point inside it experiences no net electric field because any for exterted on any point in that sphere is countered by an equal but opposite field, produced by every other point on the sphere.

    I think that's what I mean. These ideas don't really lend themselves to language do they? I have an idea, but I think it would have to be expressed visually, mathematically and verbally in order to convey its exact meaning and implications to any student.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: A rod and two balloons
  1. Two Rods (Replies: 0)

  2. Two rods and a string (Replies: 10)