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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


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    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


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    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
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