Electrostatics: Moving Charge (comprehension problem)

[jackqpublic]

I already know what the answer to this problem should be, however I do not understand why that is the case.

Homework Statement

The negative charge (= -1 micro Coulombs) in the figure below goes from y = -5 to y = 5 and is made to follow the dashed line in the vicinity of two equal positive charges (= +5C). What is the work required to move the negative charge along the dashed line?

Homework Equations

V_b - V_a = W_ab/q₀
V = kq/r
F = qE
E = kq/r²
W = Fd cosine theta

The Attempt at a Solution

Below is a representation of the arrangement of the charges. The horizontal dashes (-) are just space fillers and can be ignored, the vertical ones represent the negative charge's path.

------------y=5
-------------|
-------------|
-------------|
----+5C----y=0----+5C
-------------|
-------------|
-------------|
------------y=-5

I tried solving it on my own by taking the net displacement as 10 and coming to the conclusion that there would be work. Then I checked the answer and it said there is no net work because the force changes as the charge passes the origin. I don't understand why that is and was hoping someone could explain it to me. The only thing I could think of was that as the charge approaches the origin it is moving toward the positive charges (natural attraction) and as it moves past them it is moving away from the positive charges (against attraction).

 

[kuruman]

Look at your first equation. What is V_b - V_a?
Another way of seeing it is that the electric force is attractive, but changes sign when the charge crosses the origin. Because of the symmetry, this force does positive work over the top half of the path and negative work of equal absolute value over the bottom half of the path so that the net work is zero. Your thinking along these lines is OK.

 

[tomcue]

The reason there is no net work required to move the negative charge along the dashed line is because of the principle of superposition in electrostatics. This principle states that the total force on a charge due to multiple charges is equal to the vector sum of the individual forces from each charge. In this case, as the negative charge moves along the dashed line, it experiences a force from each of the positive charges. When the negative charge is at y=0, the forces from the two positive charges cancel each other out, resulting in no net force and therefore no net work required to move the charge. This is why the force changes as the charge passes the origin, because the direction of the force changes from attraction to repulsion. This is also why the net displacement of 10 does not result in any work, because the forces are balanced out at the origin. I hope this helps clarify the concept for you.