Respuesta :
Answer:
[tex]2.25\mu C[/tex]
Explanation:
At the beginning, we have:
V = 4.0 V potential difference across the capacitor
[tex]Q=9.0 \mu C=9.0\cdot 10^{-6}C[/tex] charge stored on the capacitor
Therefore, we can calculate the capacitance of the capacitor:
[tex]C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F[/tex]
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:
[tex]Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C[/tex]
Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from [tex]9.0 \mu C[/tex] to [tex]11.25 \mu C[/tex]. Therefore, the additional charge that moved to the positive plate is
[tex]\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C[/tex]
In a uncharged capacitor when connected to 5.0 V battery, the additional charge flows to the positive plate, is [tex]2.25\rm \mu C[/tex] .
What is capacitance of capacitor?
The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference. The capacitance of capacitor can be given as,
[tex]C=\dfrac{Q}{V}[/tex]
Here, (Q) is the electric charge and (V) is the potential difference.
Given information-
The potential difference of the first battery is 4.0 V.
The charge of the first battery is 9.0 μC.
The potential difference of the second battery is 5.0 V.
The capacitance of the first battery is,
[tex]C=\dfrac{9.0\times10^{-6}}{4}\\C=2.25\times10^{-6} \rm F[/tex]
Let the charge of the second battery is (q). Thus The capacitance of the first battery is,
[tex]C=\dfrac{q}{5}\\[/tex]
As the capacitance of the capacitor remain same. Thus put the value of C in the above equation as,
[tex]2.25\times10^{-6}=\dfrac{q}{5}\\q=11.25\rm \mu C[/tex]
The additional charge flows to the positive plate is the difference of the charge flows to the positive plate and second battery to the first battery. Thus,
[tex]\Delta q=11.25-9\\\Delta q=2.25\rm \mu C[/tex]
Thus the additional charge flows to the positive plate is [tex]2.25\rm \mu C[/tex] .
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