Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Potenital engery trouble!

  1. Sep 25, 2005 #1
    Hello everyone i have the following picutre:
    And i'm having troubles with some of the questions I answered some of them and i was wondering if u could check to see if i did those right and to see if you can help me on the one i didn't do:

    Figure 24-22 shows three paths along which we can move positively charged sphere A closer to positively charged sphere B, which is held fixed in place.

    Well the rule is for part (a) Electric Potential alwasy decreases in the direction of the eletric field. But we have 2 positved charged particles. How is there an eletric field? does an E-field go from positive to negative? So what direction would the e-field even be pointing?
    Figure 24-22
    (a) Would sphere A be moved to a higher or lower electric potential?
    (b) Is the work done by our force positive, negative, or zero?
    (c) Is the work done by the electric field due to B positive, negative, or zero?
    (d) Rank the paths according to the work our force does, greatest first (use only the symbols > or =, for example 1=2>3).
    1 = 2 = 3

    Thanks! I'm also confused on (c), first i need a way to figure out what direction the e-field is going then i'll be okay maybe!
  2. jcsd
  3. Sep 25, 2005 #2


    User Avatar
    Homework Helper

    Objects cannot push on themselves, except internally
    so the E-field that is noticed by (important for) object A
    is the E-field produced by object B (alone, not the sum).

    The question author even told you in part (c):
    only pay attention to the E-field due to B !

    Objects are effected by OTHER things ...
    this is the same rule for Electric potential
    that you had for Electric Field, and Gravity Field,
    and every Force on its Free-body-diagram ...
  4. Sep 25, 2005 #3
    thanks!! so since sphere A is moving towards its electric field, its Electric potenial must be decreasing also because B is fixed, does that mean work is zero? because u need movement for work to happen right?
  5. Sep 25, 2005 #4


    User Avatar
    Homework Helper


    Object A is moving toward object B;
    A is moving toward the *source* of the E-Field.

    The E-Field is everywhere, so A can't move toward it.

    A's displacement vector points
    in the opposite direction that the E-Field points.
    Since q(A)>0 the displacement is opposite F_on_A .

    It's hard to move an object opposite
    the other Force ... it takes Energy!
    Object A's PE is increased by our hand,
    so (since q(A) >0) A is moved to higher Potential

    Work is done ON the object that is moving ,
    by the Force that is applied TO the moving object,
    so WE DON"T CARE about the Work done on B!
  6. Sep 25, 2005 #5
    Thahnks for the explanation! it helped alot, but i'm still confused on somthing...
    I understand part (a), I also understand part (d). But The answers for part (b) and (c) i don't get. The answer for (b) Is the work done by our force positive, negative, or zero? Answer: Positive.
    Okay, "Our" force, i'm assuming is the electric force on A by B. So here is my diagram.
    F1 = Force of moving the charge.
    F2 = Electric Feild from B on A.
    d = distance A is moving.

    W = FD
    W = qEdcos(x)
    W = (+1.6C)(-1N/C)(+1m)(cos(180));
    W = (+)(-)(+)(-) = +
    ohhh, i think i just explained it by asking the question, did I prove that right?

    Okay for (c)
    Is the work done by the Electric field due to B positive, neg, or 0.
    Well isn't this the same as part (b)?
    The efield on A is pushing it away, A is moving in the opposite direction of the E-field, so wouldn't it also be:
    W = (+)(-)(+)(-) = + ?

    THanks again!
  7. Sep 25, 2005 #6


    User Avatar
    Homework Helper

    Object A's PE is increased by our hand

    that's what the author refers to
    by the words "our Force".
    Our Force on A is opposite the electric force on A.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook