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If a capacitor has a capacitance of \(10.2 \mu \mathrm{F}\) and we wish to lower the potential difference across the plates by \(60.0 \mathrm{V},\) what magnitude of charge will we have to remove from each plate?

Short Answer

Expert verified
Answer: To reduce the potential difference by 60.0 V, 612 μC of charge must be removed from each plate of the capacitor.

Step by step solution

01

Calculate the Charge To Be Removed

First, let's identify the given values: Capacitance, \(C = 10.2 \mu \mathrm{F} = 10.2 \times 10^{-6} \mathrm{F}\) Potential Difference Reduction, \(\Delta V = 60.0 \mathrm{V}\) Our goal is to find the charge \(\Delta q\) that needs to be removed to lower the potential difference by \(60.0 \mathrm{V}\). Since we know the desired voltage reduction and the capacitance of the capacitor, we can use the equation \(q = C \cdot V\) to determine the charge that must be removed: \(\Delta q = C \cdot \Delta V\)
02

Substitute the Given Values and Solve

Substitute the given values into the equation to find the change in charge: \(\Delta q = (10.2 \times 10^{-6}) \cdot 60\) Calculate the result: \(\Delta q = 612 \times 10^{-6}\)
03

Express the Result in Suitable Units

Finally, express the result in microcoulombs (\(\mu \mathrm{C}\)): \(\Delta q = 612 \times 10^{-6} \mathrm{C} = 612 \mu \mathrm{C}\) So, to lower the potential difference across the plates by \(60.0 \mathrm{V}\), we need to remove \(612 \mu \mathrm{C}\) of charge from each plate of the capacitor.

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