Problem about electric potential and power transmitted down a cable

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Printed textbook excerpt: Problem 8.1 asks to calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.58, with the two conductors at potential difference V carrying current I, down one and back up the other.
Diagram of a coaxial cable with a solid inner conductor of radius a and an outer conductor of radius b along the z axis, with current I shown flowing in opposite directions in the two conductors.
Textbook page titled Chapter 8, Conservation Laws, showing Problem 8.1 with electric field, magnetic field and Poynting vector equations, a red-boxed line reading But V = integral from a to b of E dot dl = lambda over 2 pi epsilon0 times ln(b/a), so P = IV.


When to use ## V =-\int E ⋅ dl ## or ## V =\int E ⋅ dl ## ?

When i use ## V =-\int E ⋅ dl ##. i can not show that P = IV because ##\frac{ln \frac{b}{a}}{ln \frac{a}{b}} ≠ 1##

But the solution use ## V =\int E ⋅ dl ## I'm concerned about using "minus" .
 
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  • Textbook page titled Chapter 8, Conservation Laws, showing Problem 8.1 with electric field, magnetic field and Poynting vector equations, a red-boxed line reading But V = integral from a to b of E dot dl = lambda over 2 pi epsilon0 times ln(b/a), so P = IV.
    7.png
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  • Printed textbook excerpt: Problem 8.1 asks to calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.58, with the two conductors at potential difference V carrying current I, down one and back up the other.
    8.png
    7.5 KB · Views: 784
  • Diagram of a coaxial cable with a solid inner conductor of radius a and an outer conductor of radius b along the z axis, with current I shown flowing in opposite directions in the two conductors.
    9.png
    17.5 KB · Views: 733
The notation in the solutions is kind of sloppy, but it's easy enough to understand what was meant.

Assume that ##\vec E## points radially outward. Then the inner conductor is at a higher potential than the outer conductor so that ##V = V_a - V_b##. So you have
$$V = V_a - V_b = -\int_b^a \vec E \cdot d\vec l = \int_a^b \vec E \cdot d \vec l.$$
 
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