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!

Electric Charges

  1. Dec 2, 2003 #1
    Charge A is +2.5*10^-5C Charge B is -3.7*10^-7C.
    A is 25cm away from B. Point C is 10cm away from point B. What is the electric field at C? All points are in a line A--B-C.
  2. jcsd
  3. Dec 2, 2003 #2
    [tex]A: 25\mu C, B: -.37\mu C, C: ?[/tex]

    E &= \frac{F}{q}
    &= \frac{kqQ/r^2}{q}
    &= k\frac{Q}{r^2}
    &= \frac{Q}{4\pi \sigma_{0}r^2}
    Last edited: Dec 2, 2003
  4. Dec 2, 2003 #3
    I have no idea what you just wrote. What numbrs do i put in for where?
  5. Dec 2, 2003 #4
    A, B, attract each other.

    [tex]F_{AB} = k\frac{Q_1 Q_2}{r^2} = 6.796 * 10^{21}[/tex]

    [tex]E = F/q[/tex]

    Should be easy from there.

    Edit: q could be thought as a differential, a number really close to 0, but never reach 0; so:

    [tex]E = \lim_{q-->0} \frac{F}{q}[/tex]
    Last edited: Dec 2, 2003
  6. Dec 6, 2003 #5
    How do i find the charge (q) at C for [tex]E = F/q[/tex]
  7. Dec 6, 2003 #6
    How did you figue out [tex]F_{AB} = k\frac{Q_1 Q_2}{r^2} = 6.796 * 10^{21}[/tex]
    q1= 2.5*10^-5 C
    q2=-3.7*10^-7 C

    With those numbers i got -0.6795 N

    What did i do wrong?

    Also, can this problem be solved by finding electric fields separately and then adding them??
  8. Dec 6, 2003 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    You've posted 5 questions all of which involve using the same basic formulas. Show us what you have done on the problem.
  9. Dec 8, 2003 #8
    When i use the formula [tex]F_{AB} = k\frac{Q_1 Q_2}{r^2}[/tex] Do i multiply that negative sign in the equation for a negative charge?
  10. Dec 8, 2003 #9
    This is what i did Fe=[(9.0*10^9)(2.5*10^10-5)(3.7*10^-70]/(.25)^2 =1.332

    [tex]E = F/q[/tex] , but i don't know that charge at C

    Is that what i do? Or what did i do wrong

    PS. Is there some kind of software that i can get so i don't have to keep using ^ for exponents?
  11. Dec 8, 2003 #10
    If i try this [tex]E = \lim_{q-->0} \frac{F}{q}[/tex] i will get

    [tex]E = \lim_{q-->0} \frac{F(x+q)-F(x)}{q}[/tex] , i don't kow that the numbers are? I don't think i got the step before this
  12. Dec 9, 2003 #11


    User Avatar
    Staff Emeritus
    Science Advisor

    1. There is NO charge at C. The "electric field" (since you are treating it as a scalar) at a point is, by definition, the force that would be applied to a unit charge.

    2. You don't need special software. On this forum x [ s u p]2[ / s u p] (without the spaces) will give x2.

    There is a thread at the top of each forum area called
    "Announcement: Howto Make Math Symbols Update" that explains that and more.
  13. Dec 10, 2003 #12
    Would this work?

    Find the electric fields a pont A & B. Then add the two fields ??? Or would you subtract the two fields ???

    Use the formula [tex]{E} = k\frac{Q}{r^2}[/tex]

    Is the field on point A & B going left or right?
    Last edited: Dec 10, 2003
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric Charges
  1. Electric Charges (Replies: 1)

  2. Electric charge (Replies: 3)