- #1
Ibby321
- 1
- 1
- Homework Statement
- I’ve included a picture in the attempted solutions section because there is a diagram.
- Relevant Equations
- I’m not sure which is relevant but these are all the ones I can think of:
ΔU = W = Fdcosθ = Fr = (kQq)/r
ΔV = Vb - Va = Wab/q0
W = qΔV = qEr
V = kQ/r
F = qE
E = kq/r2
I know that the answer is 0 J (no NET work was done) because there is symmetry to the problem and this symmetry comes from the fact that the direction of force changes, BUT I don’t know why the force changes (I have an idea; TBD below in #4). When I did this problem I thought I could find the voltage.
q = 1μC = 10^(-6) C
Va = k(5C)/(-5m)
Vb = k(5C)/(5m)
ΔV = Vb - Va = 1.798*10^10
Wab = ΔV/q = none of the answers
Then I thought maybe it has to do with the distance the charge traveled, so I guessed 10 J since the negative charge traveled 10 units. But that was wrong too.
I actually have many questions because I don’t have much practice with this. The question doesn’t mention anything about the 2 positive charges so I’m wondering the following:
1. I’m assuming that the two big, positive charges are to be considered as stationary source charges. Is that correct?
2. Is the fact that both charges are +5C (they are like charges of the same magnitude) significant to the symmetry of this problem? The following questions are meant as examples to explain the previous question, they don’t have to be answered: For example, if both charges are positive but one is bigger than the other (like one is +5C and the other is +3C), is there still no net work? Or if they were unlike charges of equal magnitude (for example, one is +5C and the other is -5C)? Or unlike charges of unequal magnitudes?
3. There is no x-axis but it looks like the positive charges are equidistant on the x-axis (imagining an x-axis, the one on the right is at the coordinates x,0 and the one on the left is at -x,0). Is this important to the symmetry of the problem? And if it is, am I correct?
4. Based on the explanation, I feel like I can extrapolate to vertical motion of a test charge in general. If a test charge moves vertically from y = -a to y = a (or from y = a to y = -a) between two source charges, the net work will always be zero regardless of the charges (positive or negative test charge, like or unlike test charges). The reason I say that is this: opposite charges attract, like charges repel. When the negative charge moved from y = -5 to y = 0 it was moving towards the positive source charges, which is natural behavior. Work is related to EPE, so when a negative charge moves toward a positive charge, that means the EPE is increasingly negative and that means work is negative. However when the negative charge moves from y = 0 to y = 5, it moves the same distance away from the positive source charges, which is unnatural behavior and that means a positive quantity of work is required to move it away (also EPE is relevant here). Couldn’t this logic be applied to a positive test charge moving between two positive source charges? I would think the only difference is from y = -5 to y = 0 the work would be positive and from y = 0 to y = 5, work would be positive. Generally when a test charge is moving vertically (from y = a to y = -a) and the source charges are placed horizontally, it will be moving towards the charges and then away in equal distances, which means the quantity of work cancel out. I’m not sure if any of this is right and I know it’s a lot but I’m trying here.