Electric field finding third charge

In summary, the electric field at x=0 should be zero if the point charge of +4c is placed at x=-2cm and the point charge of -6c is placed at x=1cm.
  • #1
physics10189
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Homework Statement


a point charge of +5c is located at x=-2cm, and a second point charge of -6c is located at x=1cm
where should a thid charge of +4c be placed so that electric field at x=0cm is zero?


Homework Equations



E(total)=E1+E2+E3

E=k*q/(r^2)

The Attempt at a Solution



E(total)=0
E3=-E1-E2

what I did first is divide the k out so now the equation will look like this
q3/(r3^2)=-(q1/(r1^2))-(q2/r2^2))
and then solve for r3 and put a negative sign because the force is going to the right to the electric field at x=0
i got r3=-.917cm
It is saying I am getting this wrong can anyone tell me what i am doing rong
 
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  • #2
You forgot that electric field adds vectorially, not arithmetically. Think about the configuration: if you put a positive test charge at x=0, both the +5c charge and the -6c charge would push it to the right. The third charge's field has to counter the combined effect of these two charges. Try writing an equation with only the magnitudes of the electric fields; forget about +/- signs.
 
  • #3
Well the first charge is pushing to the right and the second charge is pushing to the left thus making first plus and second negative because of their displacement and the position of the eletric field at x=0
but if we do forget about the +/- then this cannot be solve
because the E3=-E1-E2
since we have negavtives
R^2 is in E3 we have to square root negatives number...if you see where i am getting at.
thus we need +5 and -6
 
  • #4
No, both charges push to the right: the one on the left repels while the one of the right attracts, so their effects combine.

Also, when I said we should ignore all +/- signs, I meant all of them. So don't use E3=-E1-E2; use E3=E1+E2. Don't use q=-3; use q=3.
 
  • #5
ok i got the answer correct but i am not understand when you mean by both charges are pushing to the right. This i what i see as.

q1=5c----->E=0<------q2=-6c

to me this diagram makes sense because q1 is positive and q2 is negative
and the other one that doesn't make sense to me is that
how does E3=E1+E2
where the standard equation was E1+E2+E3=0
and when you move E1 and E2 to the right they become negative.
E3=-E1-E2
thanks for your help though
 

FAQ: Electric field finding third charge

1. How do you calculate the electric field of a third charge?

To calculate the electric field of a third charge, you can use the formula E = kQ/r^2, where E is the electric field, k is the Coulomb's constant, Q is the charge of the third charge, and r is the distance between the third charge and the point where you want to find the electric field.

2. What is the direction of the electric field of a third charge?

The direction of the electric field of a third charge depends on the sign of the charge. If the third charge is positive, the electric field will point away from it. If the third charge is negative, the electric field will point towards it.

3. Can the electric field of a third charge be zero?

Yes, the electric field of a third charge can be zero. This will happen when the third charge has the same magnitude but opposite sign as the other two charges in the system, resulting in a net electric field of zero.

4. How does the distance between the third charge and the other two charges affect the electric field?

The electric field of a third charge is inversely proportional to the square of the distance between the third charge and the other two charges. This means that as the distance increases, the electric field decreases, and vice versa.

5. Can the electric field of a third charge be negative?

Yes, the electric field of a third charge can be negative. This will occur when the third charge is negative and the other two charges are positive, resulting in a net electric field pointing in the opposite direction of the third charge.

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