Final answer:
The capacitance of the system with +10.0 mC and -10.0 mC charges is 1.0 millifarads (mF), and if the charges are increased to +100 mC and -100 mC, the potential difference between the conductors would be 100 volts.
Step-by-step explanation:
Capacitance and Electric Fields
To determine the capacitance of the system with net charges of +10.0 mC and -10.0 mC and a potential difference of 10.0 V, we use the formula for capacitance, C = Q/V, where Q is the charge and V is the potential difference. In this case, C = (10.0 × 10-3 C) / (10.0 V) = 1.0 × 10-3 Farads or 1.0 mF.
(b) If the charges on each conductor are increased to +100 mC and -100 mC, the capacitance, C, remains constant because capacitance depends only on the geometrical arrangement of the conductors. Therefore, the new potential difference, V', will be V' = Q'/C, where Q' is the new charge. V' = (100 × 10-3 C) / (1.0 × 10-3 F) = 100 V.