Let \(S=100 \mathrm{~mm}^{2}, d=3 \mathrm{~mm}\), and \(\epsilon_{r}=12\) for a parallel-plate capacitor. (a) Calculate the capacitance. (b) After connecting a 6-V battery across the capacitor, calculate \(E, D, Q\), and the total stored electrostatic energy. (c) With the source still connected, the dielectric is carefully withdrawn from between the plates. With the dielectric gone, recalculate \(E, D, Q\), and the energy stored in the capacitor. \((d)\) If the charge and energy found in part \((c)\) are less than the values found in part \((b)\) (which you should have discovered), what became of the missing charge and energy?

To calculate the capacitance of the parallel-plate capacitor, we can use the formula: $$C = \epsilon_0 \epsilon_r \frac{A}{d}$$ where \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\) is the vacuum permittivity, \(\epsilon_r\) is the relative permittivity of the dielectric, \(A\) is the area of the plates, and \(d\) is the separation between the plates. Plugging in the given values, we get: $$C = (8.85 \times 10^{-12} \mathrm{F/m}) (12) \frac{100 \times 10^{-6} \mathrm{m^2}}{3 \times 10^{-3} \mathrm{m}} = 3.54 \times 10^{-11} \mathrm{F}$$

First, we'll calculate the electric field E using the formula: $$E = \frac{V}{d}$$ where V is the voltage across the capacitor (6V). Using this, we get: $$E = \frac{6 \mathrm{V}}{3 \times 10^{-3} \mathrm{m}} = 2000 \mathrm{V/m}$$ Next, we'll calculate the electric displacement field D using the formula: $$D = \epsilon_0 \epsilon_r E$$ Plugging in the calculated E and known constants, we get: $$D = (8.85 \times 10^{-12} \mathrm{F/m})(12)(2000 \mathrm{V/m}) = 2.13 \times 10^{-7} \mathrm{C/m^2}$$ Now, we'll calculate the charge Q stored in the capacitor using the formula: $$Q = C \times V$$ Using the calculated capacitance and known voltage, we obtain: $$Q = (3.54 \times 10^{-11} \mathrm{F})(6 \mathrm{V}) = 2.12 \times 10^{-10} \mathrm{C}$$ Finally, we will find the total stored electrostatic energy using the formula: $$W = \frac{1}{2} CV^2$$ Plugging in the known values, we get: $$W = \frac{1}{2}(3.54 \times 10^{-11} \mathrm{F})(6 \mathrm{V})^2 = 6.37 \times 10^{-10} \mathrm{J}$$

With the dielectric removed, the capacitance changes. We can calculate the new capacitance (\(C'\)) using the formula: $$C' = \epsilon_0 \frac{A}{d}$$ Plugging in the given values, we get: $$C' = (8.85 \times 10^{-12} \mathrm{F/m}) \frac{100 \times 10^{-6} \mathrm{m^2}}{3 \times 10^{-3} \mathrm{m}} = 2.95 \times 10^{-12} \mathrm{F}$$ Since the voltage across the capacitor, V = 6V, remains unchanged, the new electric field E' would be the same as before: $$E' = 2000 \mathrm{V/m}$$ For the new electric displacement field D', we can calculate it using the formula: $$D' = \epsilon_0 E'$$ Plugging in the constants, we get: $$D' = (8.85 \times 10^{-12} \mathrm{F/m})(2000 \mathrm{V/m}) = 1.77 \times 10^{-8} \mathrm{C/m^2}$$ We can now calculate the new charge Q' stored in the capacitor using the formula: $$Q' = C' \times V$$ Using the new capacitance and known voltage, we obtain: $$Q' = (2.95 \times 10^{-12} \mathrm{F})(6 \mathrm{V}) = 1.77 \times 10^{-11} \mathrm{C}$$ Finally, we will find the new stored electrostatic energy using the formula: $$W' = \frac{1}{2} C'V^2$$ Plugging in the values, we get: $$W' = \frac{1}{2}(2.95 \times 10^{-12} \mathrm{F})(6 \mathrm{V})^2 = 5.31 \times 10^{-11} \mathrm{J}$$

Comparing the calculated values from parts (b) and (c), we notice that the charge (Q) and energy (W) stored in the capacitor decrease when the dielectric is removed: - Q without dielectric (Q') = \(1.77 \times 10^{-11}\mathrm{C}\) < Q with dielectric = \(2.12 \times 10^{-10}\mathrm{C}\) - W without dielectric (W') = \(5.31 \times 10^{-11}\mathrm{J}\) < W with dielectric = \(6.37 \times 10^{-10}\mathrm{J}\) The missing charge and energy are used to do work on removing the dielectric from between the plates. When the dielectric is removed from the capacitor, the voltage remains constant, but the capacitance decreases. As a consequence, the charge stored on the capacitor's plates and the stored electrostatic energy decrease as well. The difference in charge and energy between the cases with and without the dielectric is used to overcome the force applied by the electric field on the dielectric as it is withdrawn from between the plates. This work is ultimately dissipated as heat or other forms of energy.