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natali 33 [55]
3 years ago
12

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte

ry is then disconnected and replaced with a 5.0 V battery, with the positive and negative terminals connected in the same manner as before. Part A How much additional charge flows to the positive plate? Express your answer in microcoulombs.
Physics
1 answer:
Lelechka [254]3 years ago
5 0

Answer:

2.25\mu C

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

Q=9.0 \mu C=9.0\cdot 10^{-6}C charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from 9.0 \mu C to 11.25 \mu C. Therefore, the additional charge that moved to the positive plate is

\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C

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