Question

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 battery 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.

Answer 1

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

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