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I can never do this question!!!

  • 18-10-2009 10:17pm
    #1
    Astrogeek
    Closed Accounts Posts: 195 ✭✭


    In general Electricity questions confuse me :(

    But this one annoys me so much, everyone I ask says it's so easy but I can never remember how to do it :(

    A square of side length a has charges q, 5q, -3q and 5q places at it's respective vertices. Calculate the electric field E at it's centre.

    Is the answer (1/4*pi*e0) by q/(a*sq.rt2/2) plus the same thing with each of the other charges?
    And if it is, why is it? :confused:


Comments

  • #2
    Kevster
    Closed Accounts Posts: 2,980 ✭✭✭Kevster


    You'd learn it better if you research how to do it yourself. Do'nt you have notes on this?


  • #3
    im_invisible
    Closed Accounts Posts: 375 ✭✭im_invisible


    Bill: Whats 5q and 5q?
    Ted: 10 Q
    Bill: youre welcome


  • #4
    Podge_irl
    Moderators, Sports Moderators Posts: 28,161 Mod ✭✭✭✭Podge_irl


    Astrogeek wrote: »
    In general Electricity questions confuse me :(

    But this one annoys me so much, everyone I ask says it's so easy but I can never remember how to do it :(

    A square of side length a has charges q, 5q, -3q and 5q places at it's respective vertices. Calculate the electric field E at it's centre.

    Is the answer (1/4*pi*e0) by q/(a*sq.rt2/2) plus the same thing with each of the other charges?
    And if it is, why is it? :confused:

    q/(a*sq.rt2/2) is the distance from the centre to each of the vertices, so you have that bit right. The field from each of the charges merely adds to create the overall field felt at the centre but you need to careful about how you add them- bear in mind that some of the charges are negative and that they are all in different directions from the centre. If you have two charges q, then the net field will be zero when you are equidistant from them. So decompose the problem into different directions. You should be able to hack it from there...


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