91Ó°ÊÓ

To find the potential difference between the plates, we can use the formula for capacitance: \(C = \frac{Q}{V}\) We are given the capacitance \(C=1.20 \mathrm{nF}\) and the charge \(Q=0.800 \mathrm{\mu C}\). Solve for the potential difference, \(V\): \(V = \frac{Q}{C}\) \(V = \frac{0.800 \mathrm{\mu C}}{1.20 \mathrm{nF}}\) \(V = \frac{0.800 \times 10^{-6} \mathrm{C}}{1.20 \times 10^{-9} \mathrm{F}}\) \(V = 667 \mathrm{V}\) So the potential difference between the plates is \(667 \mathrm{V}\).

When the separation between the plates is doubled, the capacitance should change as well. The capacitance of a parallel plate capacitor is given by: \(C = \frac{\epsilon_0 A}{d}\) where \(\epsilon_0\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the separation between the plates. If we double the separation (\(d \rightarrow 2d\)), we can write the new capacitance as: \(C' = \frac{\epsilon_0 A}{2d}\) We can relate the new capacitance to the old capacitance by dividing the two equations: \(\frac{C'}{C} = \frac{\frac{\epsilon_0 A}{2d}}{\frac{\epsilon_0 A}{d}} = \frac{1}{2}\) This means that the new capacitance, \(C'\), is half of the initial capacitance: \(C' = \frac{1}{2}C\) Now we can use the same potential difference formula as before to find the new potential difference, \(V'\): \(V' = \frac{Q}{C'}\) Since \(C' = \frac{1}{2}C\), we can write: \(V' = \frac{Q}{\frac{1}{2}C}\) \(V' = 2\frac{Q}{C}\) We already know the initial potential difference, \(V\), from Part (a). So the new potential difference is twice the initial value: \(V' = 2V\) \(V' = 2 \times 667 \mathrm{V} = 1334 \mathrm{V}\) When the plate separation is doubled, the potential difference between the plates increases to \(1334\,\mathrm{V}\).