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Homework Help: Field strength

  1. Feb 9, 2008 #1
    1. The problem statement, all variables and given/known data

    q=4.0x10^-6C, 8.0x10^-6C
    d=2m

    2. Relevant equations


    Two charges of +4.0x10^-6 C and +8.0x10^-6C are placed 2.0m apart. What is the field strength halfway between them?

    3. The attempt at a solution

    netEa= (9.0x10^9)(4.0x10^-6C) + (9.0x10^9)(8.0x10^-6C)
    (2m)^2 (2m)^2
     
  2. jcsd
  3. Feb 9, 2008 #2
    I assume by field strength, they want force, i.e. coulombs law.
    Remember they want the strength at a point half way between them, half way being 1m from each of them. Use coulombs law for each charge with radius of 1m and then add the 2 forces together.

    Chris
     
  4. Feb 10, 2008 #3
    When you said with a radius of 1m you meant the (2m)^2 both become (1m)^2,right?

    The answer in the text says 3.6x10^4N/C toward smaller charge but I got 1.08x10^5N/C if I used the(1m)^2 in my calculations......

    netEa= (9.0x10^9)(4.0x10^-6C) + (9.0x10^9)(8.0x10^-6C)

    (2m)^2 (2m)^2
     
  5. Feb 10, 2008 #4
    Im not sure how you got that answer. 36000 is correct. Although I should correct myself by saying that you need to subtract the 2 forces because they are both positive, therefore oppose eachother.

    So, ((9x10^9)(4x10^-6))/1m^2 = 36000
    ((9x10^9)(8x10^-6))/1m^2 = 72000

    The force of q2 is canceling out the force of q1. So 72000-36000 = 36000 in the direction of q1.
    Make sense?

    Chris
     
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