# On a silly electric field question.

1. Apr 30, 2012

### physikamateur

1. The problem statement, all variables and given/known data
This question is actually the simplification of a more complicated one. A particle of charge +q is placed at a point A while another particle of charge +2q is placed at B, a distance of 2r to the right of A. The midpoint between the distance is denoted as O. What is the electric field intensity at O ?

2. Relevant equations
E=F/q

3. The attempt at a solution
Honestly, if particle A has a charge of +q and if B has a charge of +q, then the electric field at O ought to be zero as they cancel each other out. But I can't understand it mathematically. Suppose I consider the electric field at point O due to the particle at A, then it's electric field E(A)=(kq)/(r^2) and the electric field at point O due to particle B E(B)=(k2q)/(r^2) where k=(1/4∏ε0) Hence the total electric field E(T)=E(A)+E(B)=(3kq)/(r^2) ??? Isn't it supposedly zero ? One might say that I ought to subtract them, but my textbook says that the electric field due to multiple charges is the vector sum of the individual electric fields. (Like the principle of superposition). Please advise on my mistake. Thank you.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 30, 2012

### physikamateur

Correction: What I meant to say was "Honestly, if particle A has a charge of +q and if B has a charge of -q, then the electric field at O ought to be zero as they cancel each other out."

3. Apr 30, 2012

### SammyS

Staff Emeritus

You have only considered the magnitude of the electrics field. The electric field due to one of the charges is in a direction directly opposite of the field due to the other, so the magnitudes subtract as the vectors add.

4. Apr 30, 2012

### physikamateur

Understood. But in finding the electric field at O while working the equations, do I consider a ficticious charge to be present at O say q1 so that E(A)=[((k)(q)(q1))/(r^2)]/(q1) and the q1 cancels off to give E(A)=(kq)/(r^2) ? Moreover pertaining to the initial question, the electric field at O (E) due to +2q and +q is: E=(3kq)/(r^2) ?? Since both are positively charged and the addition of the vectors does not subtract ??

5. May 2, 2012

### elfinitty

No.You consider that there is a +q at O.So the one has a charge of +q one would push it away(eh,do you call it like that in English), whereas the one with the charge of -q would pull it towards itself.So both the vectors are the same way.You take the sum of it.