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Calculating electric field at a point due to 2 charges 
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#1
Jun2705, 03:08 PM

P: 1,629

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? Thanks 


#2
Jun2705, 03:12 PM

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P: 11,928

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. 


#3
Jun2705, 03:16 PM

Mentor
P: 41,489

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 xaxis?) 


#4
Jun2705, 03:17 PM

P: 1,017

Calculating electric field at a point due to 2 charges
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. 


#5
Jun2705, 03:28 PM

P: 1,629

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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