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Coloumb's Law problem (calculating Electrical force on a charge)

  1. Feb 3, 2010 #1
    1st Problem

    Two positive charges, each 4.18C, and a negative charge, -6.36C are fixed at the vertices of an equilateral triangle of side 0.13m.
    Find the electrical force on the negative charge.

    I don't know how to calculate electrical force on a specific charge :-S
    and please explain why we take negative charge always as positive ?

    please explain in detail why we take negative charge always as positive when applying coulomb's law.

    2nd problem:

    two charges q1 and q2, held a fixed distance d apart. (a) Find the strength of electric force that acts on qi. Assume that q1=q2=21.3C and distance=1.52m
    A third charge q3=21.3C is brought in and placed as shown in figure, find the electric force on q1 now.


    i've solved part (a) of the 2nd problem but i don't know how to solve the second part of it :-S
    please explain

    ( I don't want the answers of these problems , i just need some help and explanation )

    Attached Files:

    Last edited: Feb 3, 2010
  2. jcsd
  3. Feb 3, 2010 #2


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    Welcome to PF!

    Hi XuFyaN! Welcome to PF! :smile:
    If E is the field and q is the charge, the the force is F = qE.
    I don't understand what you mean. :confused:
    Find the force on q1 as if q3 wasn't there, and the force on q1 as if q2 wasn't there.

    That will give you two vectors, so add those vectors (either by adding the components, or by using a vector triangle ). :smile:
  4. Feb 4, 2010 #3
    Re: Welcome to PF!

    for Problem 1:

    no E is given there :-S , how can i solve it ?
    first i applied coloumb's law on two charges q1 and q2 , what should be the next step ?

    for problem 2

    i have solved (a) part by using F=kq1q2/r2
    and got the answer 1.76N , but for part (b),
    i don't know how to find the strength of the electric force on q1 charge ?


    part (b) of 2nd problem has been solved but i don't know its correct or incorrect , please have a look
    F1= F12+F13sin45

    is it correct or incorrect ?
    Last edited: Feb 4, 2010
  5. Feb 6, 2010 #4
    please reply
  6. Feb 6, 2010 #5


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    (is it 45º? the original question was about an equilateral triangle (3 x 60º), is this the same?)

    No, you've assumed the resultant force is along one of the sides, it isn't.
  7. Feb 7, 2010 #6
    so should i have to multiply sin60 by 3 ??
    bec0z according to b0ok the answer is 3.02N and i am not getting this correct :(
  8. Feb 7, 2010 #7


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    Stop guessing! :rolleyes:

    Which direction do you think the total force is in?
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