Calculating electric field at a point due to 2 charges

  1. Hello everyone, I'm just starting this problem and i'm already confused. I'm suppose to calculat the electric feild at the point P(x =3, y = 3) due to two charges, q1 = 4.0x10^-6C at the origin, and q2 = -3.0x10^-6C at x = 3.0m.

    I'll try and draw the diagram.

    ....................................^ E1
    .................................../
    ................................../
    ................................./
    ...............................(P)
    ................................|
    ................................|
    ................................v E2



    (q1).........................(q2)



    How did they know E2 will be point downard and E1 would be diagnoal like that? :bugeye: Thanks
     
  2. jcsd
  3. dextercioby

    dextercioby 12,317
    Science Advisor
    Homework Helper

    Decompose each vector in the two perpendicular components and add those 2 vectors using components.At the end,you'll have to use Pythagora's theorem to find the modulus,knowing the components.

    Daniel.
     
  4. Doc Al

    Staff: Mentor

    The charge q2 is negative and directly below the point P; so the field E2 points straight down (towards q2).

    Similarly, the field (E1) from q1, a positive charge, points away from q1 towards P, thus is at a diagonal. (What angle does it make with the x-axis?)
     
  5. I don't know, but you can just add the two electric fields together separately. The electric field at distance r is:

    E(r) = Q/(4*pi*e0*r^2), where e0 is the permittivity of free space (8.854*10^-12).

    At P, r from the charge q1 is 4.24 (root of 2*3^2), while r at P from q2 is 3. Calculate these two fields, add 'em together. Remember that the two fields aren't pointing in the same direction though - the y part of the field due to q1 can be added to the field due to q2, but not the x part.
     
  6. ahh i remember now! Sorry its been a year since i took this course and i forgot literally everything. It will make a 45 degree angle. I ended up getting the answer! thanks!
     
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