Short webpage title: Calculating Electric Field and Force from Two Charges

[dgl7]

Two charges create an electric field--electric field strenght at a point above h.fiel

Homework Statement

Two charges are located on a horizontal axis. The Coulomb constant is 8.98755x10^9 Nm^2/C^2.

a) Determine the electric field at p on a vertical axis as shown in the attachment. Up is the positive direction. Answer in units of V/m.

b) Calculate the vertical component of the electric force on a -3.1e-6C charge placed at point p. Answer in units of N

Homework Equations

a) E=kQ/r^2
b) F=Eq

The Attempt at a Solution

a) I'm fairly certain I know how to find the field strength from the two charged particles, if point p was directly on the field and in the center.

E=E1+E2
E=k/r^2(Q+q)
E=8.98755e9/3^2(2.2e-6+2.2e-6)
E=4393 V/m

But I think I also need to take into account that p is above the two particles, but now I'm stuck...

b) I think this part would be easier, I just can't do it since I haven't gotten the answer to part a.

F=Eq
F=E(that would be found in part a)(-3.1e-6)

 

[tiny-tim]

Hi dgl7!

E=k/r^2(Q+q)
E=8.98755e9/3^2(2.2e-6+2.2e-6)

That's correct, if p was on the x-axis, you would use r = 3 for each charge, and add.

Instead, use r = the actual distance between p and each charge.

That will give you the force, so remember force is a vector, and add the two vectors.

 

[dgl7]

AHHH that makes sense. Thanks very much!

 

[dgl7]

Ok nope. Nevermind, I'm still confused.
This is what I've been doing and attempting:
E=Eleft+Eright
E=kQleft/r^2+kQright/r^2
(r=r and Qleft=Qright)
E=2kQ/r^2
E=2*8.98755e9*2.2e-6/(1.8^2+3^2)
E=3230.818627 V/m

Not sure what I'm doing wrong.

 

[tiny-tim]

The electric field is a vector,

so the electric field from each charge has the magnitude kQ/r², but it also has a direction.

The direction is different for each charge, so you can't just add the magnitudes, can you?