What is the electric field intensity at point A due to two point charges?

AI Thread Summary
The discussion focuses on calculating the electric field intensity at point A due to two point charges, q1 and q2, with given values. The electric field created by each charge is determined using the formula E = kq/r^2, resulting in E1 = 2.8x10^9 N/C and E2 = 1.9x10^10 N/C. A vector diagram is created to visualize the direction of the electric fields, which point away from positive charges. The need for an angle to use the component method of vector addition is highlighted, with clarification that the angle is not 90 degrees but should be determined through trigonometry. Accurate values for distance from each charge to point A are emphasized for correct calculations.
Inertialforce
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Homework Statement


Find Enet (magnitude and direction) at point A and include a vector diagram.

q1 = 2.0x10^-6 C
q2 = 6.0x10^-6 C
r1 = 2.5x10^-3 m
r2 = 1.7x10^-3 m

Homework Equations


E = Klql/r^2 or E = Kq/r^2 (the first equation just has absolute values around the q for reminder but they both are the same thing)


The Attempt at a Solution


First I looked at the diagram and I know that the electric field created by a charge will always put away from a positive charge and towards a negative so I drew a vector diagram for that at point "A" using tip to tail method, then I joined E1(electric field created by q1) and E2(electric field created by q2) to have a resultant of E.

I then used the equation E = kq/r^2 to find out what E1 and E2 are and I found out that they equal 2.8x10^9 and 1.9x10^10 respectively. My question, now I have to use the component method of vector addition to find out what E is but I need an angle to do that. Is the angle in this case 90 degrees?
 

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Inertialforce said:
I then used the equation E = kq/r^2 to find out what E1 and E2 are and I found out that they equal 2.8x10^9 and 1.9x10^10 respectively.
How did you find E2? What value of r (or r^2) did you use?
My question, now I have to use the component method of vector addition to find out what E is but I need an angle to do that. Is the angle in this case 90 degrees?
No. Figure out the angle that E2 makes with the horizontal by examining that triangle and using a little trig.
 
Doc Al said:
How did you find E2? What value of r (or r^2) did you use?

I figured out E2 by using the values q2 = 6.0x10^-6C and r(or r^2) = 1.7x10^-3m. Because these values were given at the start of the question, we didn't need to find them.

Then I took those numbers and I directly plugged it into the equation E= kq/r^2.
 
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Inertialforce said:
I figured out E2 by using the values q2 = 6.0x10^-6C and r(or r^2) = 1.7x10^-3m. Because these values were given at the start of the question, we didn't need to find them.

Then I took those numbers and I directly plugged it into the equation E= kq/r^2.
To figure out E2 (the field from q2 at point A) you need the distance from q2 to point A, not the distance between the charges (which is given as 1.7x10^-3m).
 
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