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Two charges create an electric fieldelectric field strenght at a point above h.fiel 
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#1
Jan2610, 09:57 AM

P: 8

1. The problem statement, all variables and given/known data
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 3.1e6C charge placed at point p. Answer in units of N 2. Relevant equations a) E=kQ/r^2 b) F=Eq 3. 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(2.2e6+2.2e6) 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)(3.1e6) 


#2
Jan2610, 02:59 PM

Sci Advisor
HW Helper
Thanks
P: 26,148

Hi dgl7!
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. 


#3
Jan2610, 04:10 PM

P: 8

AHHH that makes sense. Thanks very much!



#4
Jan2610, 05:27 PM

P: 8

Two charges create an electric fieldelectric field strenght at a point above h.fiel
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*2.2e6/(1.8^2+3^2) E=3230.818627 V/m Not sure what I'm doing wrong. 


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