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Answer :
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Explanation:
1.55 V is the answer
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The calculated voltage is 1.55 V. To calculate the voltage applied to a 2.00 μF capacitor holding 3.10 μC of charge, use the formula V = Q / C.
In Physics, the relationship between the charge (Q), capacitance (C), and voltage (V) of a capacitor is given by the formula:
[tex]Q = C \times V[/tex]
We can rearrange this formula to solve for voltage (V):
[tex]V = Q / C[/tex]
Given the values:
- Q = 3.10 μC
- C = 2.00 μF
We convert the units to their base SI units:
- [tex]3.10\left \mu C = 3.10 \times 10^{-6}\left C[/tex]
- [tex]2.00\left \mu F = 2.00 \times 10^{-6}\left F[/tex]
Substituting the values into the formula:
[tex]V = 3.10 \times 10^{-6} C / 2.00 \times 10^{-6} F = 1.55 V[/tex]
Therefore, the voltage applied to a 2.00 μF capacitor when it holds 3.10 μC of charge is 1.55 V.
Complete Question:
Calculate the voltage applied to a 2.00 μF capacitor when it holds 3.10 μC of charge.
