A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm},\) is connected to a \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

We will use the formula for the capacitance of a parallel plate capacitor: \(C = \frac{\epsilon_0A}{d}\) where \(C\) is capacitance, \(\epsilon_0\) is the vacuum permittivity, \(A\) is the plate area, and \(d\) is the separation between the plates. Given values: \(A = 12.0 \mathrm{~cm}^{2} = 1.2\times10^{-3} \mathrm{~m}^{2}\) \(d_{1} = 1.50 \mathrm{~mm} = 1.50\times10^{-3} \mathrm{~m}\) So, the initial capacitance is: \(C = \frac{(8.854\times10^{-12} \mathrm{~F/m})\times(1.2\times10^{-3} \mathrm{~m}^{2})}{1.50\times10^{-3} \mathrm{~m}} = 7.084\times10^{-12} \mathrm{~F}\)

With the initial capacitance, we can find the initial charge stored on the capacitor using the formula: \(Q = CV\) where \(Q\) is charge, \(C\) is capacitance, and \(V\) is voltage. Given voltage, \(V = 9.00\; \mathrm{V}\), we can compute the initial charge: \(Q = (7.084\times10^{-12}\;\mathrm{F})(9.00\;\mathrm{V}) = 63.76\times10^{-12}\;\mathrm{C}\)

The charge on a capacitor is conserved when it is disconnected from any voltage source. Thus, as the plates are pulled apart, the charge remains the same. We can calculate the final capacitance and voltage as the separation distance increases. Given final separation distance: \(d_{2} = 2.75 \mathrm{~mm} = 2.75\times10^{-3} \mathrm{~m}\) The final capacitance is: \(C_{2} = \frac{(8.854\times10^{-12} \mathrm{~F/m})\times(1.2\times10^{-3} \mathrm{~m}^{2})}{2.75\times10^{-3} \mathrm{~m}} = 3.863\times10^{-12} \mathrm{~F}\) The final voltage across the capacitor when the separation increases is: \(V_{2} = \frac{Q}{C_{2}}\) \(V_{2} = \frac{63.76\times10^{-12}\;\mathrm{C}}{3.863\times10^{-12} \mathrm{~F}} = 16.51\; \mathrm{V}\)

Now, we will calculate the change in the energy stored in the capacitor and find the work done. The energy stored in a capacitor is: \(E=\frac{1}{2}CV^2\) The initial energy stored in the capacitor is: \(E_{1}=\frac{1}{2}(7.084\times10^{-12}\;\mathrm{F})(9.00\;\mathrm{V})^2 = 2.874\times10^{-9}\;\mathrm{J}\) The final energy stored in the capacitor after separation increases is: \(E_{2}=\frac{1}{2}(3.863\times10^{-12}\;\mathrm{F})(16.51\;\mathrm{V})^2 = 2.113\times10^{-9}\;\mathrm{J}\) Since the battery is connected to the capacitor, we must consider the energy change in the battery as well. The work done on the plate can be expressed as: \(W = E_2 - E_1\) \(W = 2.113\times10^{-9}\;\mathrm{J} - 2.874\times10^{-9}\;\mathrm{J} = -7.61\times10^{-10}\;\mathrm{J}\) The result is negative since the energy used to pull again the capacitor's force is stored back into the battery due to its negative terminal. Hence, the work done in pulling the plates of the capacitor apart is \(7.61\times10^{-10}\;\mathrm{J}\).