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

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Homework Help Overview

The discussion revolves around calculating the electric field intensity at a specific point (point A) due to two point charges, q1 and q2, with given magnitudes and distances. The participants are exploring the implications of electric field direction and vector addition in the context of electrostatics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of electric fields E1 and E2 using the formula E = kq/r^2 and the implications of charge polarity on field direction. Questions arise regarding the angle needed for vector addition and the correct distance to use for calculating E2.

Discussion Status

There is an ongoing examination of the calculations for E1 and E2, with some participants seeking clarification on the values used and the geometry involved. Guidance has been offered regarding the need to determine the angle for vector addition, suggesting a focus on trigonometric relationships.

Contextual Notes

Participants are working within the constraints of the problem statement, which provides specific values for charges and distances. There is an emphasis on understanding the setup and the relationships between the charges and point A, with some confusion noted regarding the distances involved.

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.
 
Last edited by a moderator:
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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