Question

The plates of a 15.0 - \(\mu \mathrm{F}\) capacitor have net charges of $+0.75 \mu \mathrm{C}\( and \)-0.75 \mu \mathrm{C},$ respectively. (a) What is the potential difference between the plates? (b) Which plate is at the higher potential?

Short answer

Answer: The potential difference between the plates is 0.05 V, and the plate with the higher potential is the one with the +0.75 μC charge.

Step-by-step solution

Find the potential difference between the plates

Using the given information, we can use the formula for the capacitance of a capacitor, which is \(C = \frac{Q}{V}\), where \(C\) is the capacitance, \(Q\) is the charge, and \(V\) is the potential difference. We can rearrange this formula to solve for the potential difference: \(V = \frac{Q}{C}\). Insert the given values: \(C = 15.0 \, \mu \mathrm{F} = 15.0 \times 10^{-6} \mathrm{F}\) \(Q = 0.75 \, \mu \mathrm{C} = 0.75 \times 10^{-6} \mathrm{C}\) Calculate the potential difference: \(V = \frac{0.75 \times 10^{-6} \mathrm{C}}{15.0 \times 10^{-6} \mathrm{F}}\)

Calculate the potential difference

Now we can calculate the potential difference: \(V = \frac{0.75 \times 10^{-6} \mathrm{C}}{15.0 \times 10^{-6} \mathrm{F}} = 0.05 \mathrm{V}\) So the potential difference between the plates is 0.05 V.

Determine the plate with the higher potential

From the problem statement, we have the charges on the plates as +0.75 \(\mu\)C and -0.75 \(\mu\)C. The plate with the higher potential will be the plate with the positive charge, which is the plate with a charge of +0.75 \(\mu\)C. To summarize: a) The potential difference between the plates is 0.05 V. b) The plate with the higher potential is the one with the +0.75 \(\mu\)C charge.