Calculating electric field at a point due to 2 charges

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To calculate the electric field at point P due to two charges, q1 and q2, one must consider the direction of each electric field vector. The electric field E2 from the negative charge q2 points downward towards q2, while E1 from the positive charge q1 points diagonally away from q1 towards P. The magnitudes of these electric fields can be calculated using the formula E(r) = Q/(4*pi*e0*r^2), with distances from each charge to point P being 4.24m for q1 and 3m for q2. The components of the electric fields must be added separately, taking into account their directions, particularly that E1 makes a 45-degree angle with the x-axis. The calculations ultimately lead to the correct answer for the electric field at point P.
mr_coffee
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Hello everyone, I'm just starting this problem and I'm already confused. I'm suppose to calculat the electric field 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
 
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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.
 
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?)
 
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.
 
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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