# Homework Help: Physics 2 Electric Fields

1. Sep 2, 2010

### Jgs19

Hey I'm a new time member but I hope to use this forum for my college Physics 2 class, anyways:

1. Question: Two charges are D = 2.5 m apart and lie on the x-axis. What must be the magnitude of the charge on the right one, Q2, (in C) so that the electric field is zero at x = 1.75 meters. Let Q1 be at the origin and have a charge 17 C.

2. Equations I'm using.
E(total)=E1+E2...
E=k*q/r^2

3. My attempt:
E1=(8.99E9)(17C) / (1.75^2) = 4.990E10
E2= k*Q / (1.75^2)
E(total) = E1 + E2
0 = E1+E2
-E1=E2
Q= (-4.990E10)(1.75^2) / (8.99E9)
Q= -16.9987

I've just learned electrical fields / coulombs law, for a problem like this I just need some explanation on what I'm doing wrong. I thing what the biggest problem is and confusing this is it initially is 2.5m apart but they want you to find it at 1.75m, I'm going to keep working but I feel like the 2.5 is relevant and I'm not using it. Thanks in advance.

2. Sep 3, 2010

### Socket93

it's quite simple if you understand the concepts.....see electric field is a vector, so you should take the direction under consideration before setting the equation,According to data,Q1=+17C, so the field due to it on point(1.75,0) is towards +ve x-axis.So to nulify its effect and make field zero at (1.75,0),Q2 would be -ve and direction of field due to it would be towards -ve x-axis.Now if we only talk about the magnitude,then,.....
E1=E2
so, Q1/r1^2=Q2/r2^2
so, 17/(1.75)^2=Q2/(1.25)^2
so, Q2=8.673C.
This was the magnitude that we found.Now, as we discussed earlier, Q2 has to be -ve.So,
Q2=-8.673C.