Final answer:
The potential difference across a 10µF capacitor, initially charged to 50V and then given an additional 200µC charge, will be 70V. This is calculated using the relation V=Q/C, by first finding the total charge and then the new voltage.
Step-by-step explanation:
The question involves a capacitor initially charged with a 50V battery and then disconnected, after which an additional 200µC of charge is given to the capacitor. The concept here is that the potential difference (V) across the capacitor is related to the charge (Q) on the capacitor and its capacitance (C) by the formula V = Q/C. Initially, the charge on the capacitor can be calculated using Q = CV, where C is 10µF and V is 50V. After the battery is disconnected and additional charge is given, the total charge on the capacitor changes, but the capacitance remains the same, allowing us to find the new potential difference.
First we calculate the initial charge: Q = CV = (10µF)(50V) = 500µC. With an additional charge of 200µC, the total charge becomes 700µC. Now we can find the new potential difference across the capacitor: V = Q/C = 700µC / 10µF = 70V.
The potential difference across the capacitor will be 70V, which is not one of the options provided in the question. Therefore, it might be a trick question or the options given contain an error.