Answer
The magnitude of the potential difference is $~~\frac{19}{648}~A$
Work Step by Step
We can use Equation (24-18) to find the potential difference between $r = 2.00$ and $r = 3.00$:
$\Delta V = -\int_{2.00}^{3.00}~E~dr$
$\Delta V = -\int_{2.00}^{3.00}~\frac{A}{r^4}~dr$
$\Delta V = -(-\frac{A}{3r^3})\Big\vert_{2.00}^{3.00}$
$\Delta V = \frac{A}{(3)(3.00)^3}- \frac{A}{(3)(2.00)^3}$
$\Delta V = \frac{A}{81}- \frac{A}{24}$
$\Delta V = \frac{8A}{648}- \frac{27A}{648}$
$\Delta V = -\frac{19}{648}~A$
The magnitude of the potential difference is $~~\frac{19}{648}~A$
