Electrostatics: Understanding this "Work Done" Line Integral Question

[Master1022]

There may be scope for ambiguity here. There's keeping the direction of the vector $d\vec r$ constant, and there's keeping constant the direction defined as positive for it.

Viewing the integral as an ordered sum, each $d \vec r$ is a step along the path from the lower bound to the upper bound. If the direction from R to ∞ is defined as positive, and R is the lower bound, then each $d \vec r$ is a positive step. If we reverse the bounds then we are stepping from ∞ to R, so each vector element is a negative step.Following my reasoning above, setting the lower bound as R and the upper as infinity determines that $d \vec r$ is outwards.

Okay, I think this makes more sense to me now. To see if I am understanding what you mean: I should keep the d \vec r vector pointing in the same way as the parameter r increases (i.e. outwards)?

If you are using $\vec r$ as the vector from the origin, then this is different from using $d \vec r$ as the inwards line element.

Swapping between them could be the root of your problem.

Yes, thinking about this a bit more, that does seem to be the problem.

Is this also true for general line integrals? For example, if we want to integrate along a path, parameterized by p, then we ought to have dp in the direction of (or tangential to the path at that point) of p increasing regardless of which direction we want to go (from high to low p or vice versa)?

 

[kuruman]

All this confusion can be avoided by following this procedure when doing line integrals.
1. Write the field vector in the coordinate system of your choice using unit vector notation.
Here, $\vec E=\frac{kQ}{r^2}~\hat r$
2. Write the element $d\vec l$ in standard unit vector notation ignoring the path direction. The convention is that $d(something)$ is positive when the $(something)$ is increasing.
Here, $d\vec l=dr~\hat r$. In Cartesian coordinates it will always be $d\vec l=dx~\hat x+dy~\hat y+dz~\hat z$.
3. Take the the dot product as suggested by the integrand.
Here, $\vec E \cdot d\vec l=\frac{kQ}{r^2}~\hat r \cdot (dr~\hat r)=\frac{kQ}{r^2}dr$.
4. Integrate using the starting point as the lower limit and the end point as the upper limit. The limits define the sign of the integral.

If you follow this procedure, you cannot go wrong because it's based on the formal definition of the inner product and not on the $AB\cos\theta$ thing.