Is the Work Done by the Electric Field Positive, Negative, or Zero?

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The discussion revolves around the work done by an electric field on a test charge moving between two negatively charged rods. The user correctly identifies that the electric forces at points X and Y are directed away from the test charge. They conclude that the work done by the electric field from the left rod is zero due to perpendicular vectors, while the right rod's contribution is positive, as the force and displacement vectors form an angle less than 90 degrees. However, confusion arises regarding the relationship between electric potential energy and the direction of force, leading to questions about whether the work is ultimately positive or negative. Clarification on these concepts is sought to resolve the conflicting interpretations.
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Homework Statement



Two charged rods, each with a net charge -Q are held in place as shown in the top view diagram.

a) A small test charge -q travels from point X to pt. Y along a circular arc shown.
i) Draw an arrow on the diagram at each pt. (X & Y) to show the direction of the electric force on the test charge at that point​

b.) Is the work done on the charge by the electric field positive, negative, or zero? Explain

Diagram --> http://img136.imageshack.us/img136/2262/physics2ai4.png

Homework Equations


F = q(source)E
W = Fd cos (theta)
W = q(source) Ed cos(theta)


The Attempt at a Solution


Was wondering if my answers made sense. Thanks again

a.) My reasoning for the direction is that each rod exerts an electrostatic force on the charge at points X and Y that are directed away from the charge. After drawing respective components for each force, I arrived at my present answer (in the diagram link)

b.) Here is where I'm at odds. I figured that the work due to the electric field from the left rod is zero since the electric field vector and displacement vector are perpendicular at every point along the path. Therefore the net work is only due to the right hand rod. I think the work done is positive since the electric field vectors from the right hand rod and the displacement vectors form an angle greater than 90 since the electric field is directed inwards toward the rod. However since the electrostatic force is repulsive between the right hand rod and test charge, then the electrostatic force is in the opposite direction of electric field. Thus the electrostatic force vector and displacement vector form an angle less than 90 and therefore positive work is done by the electric field from second equation. Is my reasoning logical or is there something I'm missing. Thanks
 

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Hmmm.. my textbook says that if a negatively charge particle accelerates in a direction opposite to the electric field, the charge field system looses electric potential and electric potential energy is positive. If that's true, then by the conservation of energy, kinetic energy decreases and thus work is negative? I'm really confused now... anyone?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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