Electric fields and forces help

In summary, there are 4 equal charges of 2 micro Coulombs arranged in a square shape, with 2 positives on one side and 2 negatives on the other. The question asks for the magnitude of the electric field at the center, and the answer is zero. However, if a -4 micro Coulomb charge is placed at the center, it will experience a repulsive force in the positive x-direction due to the two negative charges and an attractive force from the two positive charges. This is similar to a double electric dipole.
  • #1
mindhater
6
0
There are 4 equal charges of 2 micro Coulumbs in a shape of a square. 2 positives line up on 1 side, which is the left side and 2 negatives line up on the other side, which is the left. Again, the charges are of equal magnitude.

The question asks for the magnitude of the electric field at the center...

I was thinking it's zero, but I'm not quite sure if that's true.

The other question is if you put a -4 micro Coulumb charge at the center, what will be the magnitude force on it and direction?

I know the 2 positive charges are attracted to the negative charge, therefore there are two vector arrows coming out of the postive charges into the -4 micro C charge, but I don't which direction the arrow goes when it comes to the 2 negatives. Is there a arrow that comes out of the -4 charge towards the negative charage, or an arrow from the negative charge to the -4 charge. Since it's repulsive i don't know which one.


Any help is appreciated...thx
 
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  • #2
Since nothing is said about the size of the square, let's assume that it has side 2 and so distance to the center from each vertex *radic;(2). Set up a coordinate system so that the four corners are (1,1), (1,-1), (-1,1), and (-1,-1) with the two positive charges at the first two points. Assuming a "+1" test charge at the center, the force due to the point at (1,1) has magnitude 2/2= 1 and has components (1/&radic(2))(-i- j) (Since it will "push" a positive test charge. Similarly, the force due to the charge at (1,-1) is (1/&radic(2))(-i+ j) while the forces due to the charges at (-1,1) and (-1,-1) are (1/&radic(2))(-i- j) and (1/&radic(2))(-i+ j) respectively (Since they will "pull" the test charge toward them). The total force on a test charge, and so the field strength, will be the sum: (1/&radic(2))((-1-1-1-1)i+(1-1+1-1)j)= -4i.

Putting a charge of -4 in the center gives a force of (-4)(-4i)= 16i. That is, the two negative charges repel while the two positive charges attract and it moves in the positive x-direction.

By the way, thinking "I know the 2 positive charges are attracted to the negative charge" is misleading. The two positive charges are fixed in place and aren't moving. It is the negative charge in the middle that is being attracted. That could cause you to have your vectors reversed.
 
  • #3
It's kind of like a double electric dipole... isn't it?
 

1. What is an electric field?

An electric field is a region in space around an electrically charged object where other charged particles experience a force.

2. How is an electric field created?

An electric field is created by a charged object, such as a proton or electron. The electric field is strongest closer to the charged object and decreases in strength as you move further away.

3. What is the difference between electric field and electric force?

An electric field is a property of space that affects charged particles, while electric force is the actual force that is exerted on a charged particle within an electric field.

4. How can electric fields and forces be used in everyday life?

Electric fields and forces are used in a variety of everyday devices, such as electric motors, generators, and household appliances. They are also used in medical equipment, such as MRI machines, and in telecommunications technology.

5. What is the relationship between electric fields and potential energy?

Electric fields and potential energy are closely related. Electric potential energy is the potential energy a charged object has due to its position in an electric field. The strength of the electric field determines the amount of potential energy a charged object possesses.

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