# Electrical fields question

## Homework Statement

Two very large sheets of charge are seperated by a distance d. One sheet has a surface charge density +o and the other a surface charge density -o. A small region near the center of the sheets is shown.
1. Draw arrows on the diagram to indicate the direction of the electric fields at points A, B, C, and D
2. Compare the magnitudes of the electric fields at points A, B, C and D
3. How would the electric force exerted on a charged particle at point A compare to the electric force exerted on the same particle at point B? point C? point D?

E = F/q

## The Attempt at a Solution

1. All pointing to the right
2. B>A=C>D. Electric field is imagining putting +1C of charge; therefore it would repel the + charge and attract to the - charge
3. I have trouble on this question. I think that at points A and C, E-force would be 0 because the electric force from + and - repel. At point B and D, it would be the same because they are equidistant from the electrodes?

#### Attachments

• TUT 3.bmp
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ranger
Gold Member
It does not matter where are charge is placed in between the plates. The net force is always the same becuase the electric field is uniform between two parallel plates. If we drop a negative test charge in, the negative plate will repel it, while the positive plate attracts it. Even if its in the middle, it will be repelled by one plate and pulled by the other.

I thought electric field is always putting +1C of charge at the places on the points. Therefore, the closer it is to the positive side, the stronger the electric field, and the faster it would get repelled by the positive charge

ranger
Gold Member
Yea, youre right. Its the force experienced by placing a test charge at a point in space. But it seems like your diagram and the way its described would come out to be a uniform electric field. So no matter where the test charge, the force will always be the same.

Sorry, but I am just not getting this. E = k (q/r^2). As you can see, E depends on the distance between the charges acting on it. I know you're definitely right because the book says so, but it really did not explain much into it.

oh okay, i just got it from one of your replies ranger, thanks