# Electric Potential and bounds of integration

1. Mar 27, 2013

### timmastny

1. The problem statement, all variables and given/known data

Find the potential difference between V(P) - V(R) and V(c) - V(a)

2. Relevant equations

Electric potential:

V(a) - V(b) = $\int^{b}_{a}$E*dr

V(b) - v(a) = -$\int^{a}_{b}$E*dr

Fundamental Theorem of Calculus:

F(b) - F(a) = $\int^{b}_{a}$f(x)dx

3. The attempt at a solution

My question is about the solutions attached. In the first example, we have V(P)-V(R), which makes the integral $\int^{R}_{P}$E*dr. However, when evaluating the integral, the solution takes V(P)-V(R) which seems to disagree with the fundamental theorem of calculus. As per the bounds, I think it should be V(R)-V(P).

Note: while I think it is wrong, the online homework said that evaluting it V(R)-V(P) is incorrect.

Likewise, to add to the confusion, the second picture is the solution for V(c)-V(a), which makes an integral $\int^{a}_{c}$E*dr. In this case, though, the solution is found by taking v(a)-v(c). (technically v(a) - v(b) because v(c)-v(b) is a constant). This result seems to follow the fundamental theorem of calculus and is the result I expected.

Finally, I thought maybe it had to do with the geometry so I attached a picture of the scenario. Hopefully some one can shed some light on the confusing integral bounds. Thanks

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• ###### electpot.png
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2. Mar 27, 2013

### rude man

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No. V(b) - V(a) = -∫E*dr with lower limit of r(a) and upper limit of r(b), with * denoting the vector dot-product. If a unit test charge is moved from a to b where r(a) > r(b) then this integral is positive and represents both the gain in potential and the work done in moving a unit test charge from a to b.

OK, yor formula is not wrong quantitatively but you should think of going from a to b as integrating from a to b, and that means V = -Edr.

Last edited: Mar 27, 2013
3. Mar 27, 2013

### rude man

This is a contradiction in terms!