On a silly electric field question.

In summary, the question asks for the electric field intensity at point O, which is the midpoint between two particles of charge +q and +2q placed at points A and B respectively. The electric field at O is found by taking the vector sum of the individual electric fields due to each particle, using the equation E=F/q. The magnitude of the electric field due to each particle is given by E(A)=(kq)/(r^2) for particle A and E(B)=(k2q)/(r^2) for particle B, where k=(1/4∏ε0). However, since the direction of the electric field due to each particle is opposite, the vectors add instead of subtract, giving a total electric field of
  • #1
physikamateur
12
0

Homework Statement


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 ?


Homework Equations


E=F/q


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.

 
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  • #2
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
physikamateur said:

Homework Statement


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 ?

Homework Equations


E=F/q

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.

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
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
physikamateur said:
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."

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.
 

1. What is an electric field?

An electric field is a region of space surrounding an electrically charged object, in which other charged objects experience a force.

2. How is an electric field created?

An electric field is created by the presence of an electric charge. Electric charges can be positive or negative, and the electric field they create is strongest closer to the charge and weaker farther away.

3. What is the unit of measurement for electric field?

The unit of measurement for electric field is newtons per coulomb (N/C) in the SI system, or volts per meter (V/m) in the CGS system.

4. How does an electric field affect charged particles?

An electric field exerts a force on any charged particles in its vicinity. The direction of the force depends on the sign of the charge and the direction of the electric field.

5. Can an electric field be shielded?

Yes, an electric field can be shielded by surrounding the charged object with a conductive material. This material will redistribute the electric charge and create an equal but opposite electric field, effectively canceling out the original field.

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