Calculating electric field at a point due to 2 charges

In summary, the conversation discusses the calculation of the electric field at point P due to two charges, q1 and q2, located at the origin and x = 3.0m respectively. The direction of the electric fields from each charge is determined based on their respective charges, with E2 pointing downward and E1 at a diagonal angle. The conversation also mentions using Pythagoras' theorem to find the modulus of the electric field at P, and suggests adding the two electric fields together separately. The conversation concludes with the mention of the angle that E1 makes with the x-axis and the successful calculation of the answer.
  • #1
mr_coffee
1,629
1
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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  • #2
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
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?)
 
  • #4
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
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!
 

1. How do I calculate the electric field at a point due to 2 charges?

To calculate the electric field at a point due to 2 charges, you can use the equation E = kQ/r^2, where E is the electric field, k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance between the point and the charge. You will need to calculate the electric field for each charge separately and then add them together to find the total electric field at the point.

2. What is the direction of the electric field at a point due to 2 charges?

The direction of the electric field at a point due to 2 charges is determined by the direction of the force that would be exerted on a positive test charge placed at that point. The direction of the electric field will be away from positive charges and towards negative charges.

3. Can the electric field at a point due to 2 charges be negative?

Yes, the electric field at a point due to 2 charges can be negative. This would occur when the charges are of opposite sign and the electric field vectors are pointing in opposite directions.

4. How does the distance between the charges affect the electric field at a point?

The electric field at a point due to 2 charges is inversely proportional to the square of the distance between the charges. This means that as the distance between the charges increases, the electric field at a point will decrease. Similarly, as the distance decreases, the electric field at a point will increase.

5. Is the electric field at a point due to 2 charges affected by the magnitude of the charges?

Yes, the electric field at a point due to 2 charges is directly proportional to the magnitude of the charges. This means that as the charges increase in magnitude, the electric field at a point will also increase. Similarly, as the charges decrease in magnitude, the electric field at a point will decrease.

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