Final answer:
Doubling the plate separation of a parallel-plate capacitor would double the potential difference to maintain the same electric field strength, as the electric field in a capacitor is proportional to the charge on the plates. Therefore, the answer is E. none of the above.
Step-by-step explanation:
If the plate separation of an isolated charged parallel-plate capacitor is doubled, the correct effect on the capacitor's characteristics from the options provided is: the potential difference is doubled. This is because the electric field (E) in a parallel-plate capacitor is given by E = V/d, where V is the potential difference and d is the separation between the plates. When the plate separation is doubled, the electric field remains unchanged (since the charge remains the same and the electric field strength is directly proportional to the charge on the plates). As a result, the potential difference must also double to maintain the same electric field strength.
The charge on each plate does not change, and therefore, neither does the surface charge density, since it is defined as the charge per unit area (Q/A), and there is no indication that the area changes. Therefore, the correct answer is E. none of the above