Consider a parallel-plate capacitor having an area of \(3225 \mathrm{~mm}^{2}\left(5\right.\) in. \(^{2}\) ), a plate separation of \(1 \mathrm{~mm}(0.04\) in.), and a material having a dielectric constant of \(3.5\) positioned between the plates. (a) What is the capacitance of this capacitor? (b) Compute the electric field that must be applied for \(2 \times 10^{-8} \mathrm{C}\) to be stored on each plate.

To calculate the capacitance of a parallel-plate capacitor, we use the following formula: $$ C = \epsilon \cdot \kappa \cdot \frac{A}{d}, $$ where - \(C\) is the capacitance, - \(\epsilon\) is the vacuum permittivity (around \(8.85 \times 10^{-12} \mathrm{F/m}\)), - \(\kappa\) is the dielectric constant of the material between the plates, - \(A\) is the area of the plates, and - \(d\) is the distance between the plates. We are given \(\kappa = 3.5\), \(A = 3225 \mathrm{mm}^2 = 3225 \times 10^{-6} \mathrm{m}^2\), and \(d = 1 \mathrm{mm} = 1 \times 10^{-3} \mathrm{m}\). Substitute these values into the formula above to find the capacitance: $$ C = (8.85 \times 10^{-12} \mathrm{F/m}) \cdot 3.5 \cdot \frac{3225 \times 10^{-6} \mathrm{m}^2}{1 \times 10^{-3} \mathrm{m}} = 1.03 \times 10^{-10} \mathrm{F}. $$

To store a charge of \(Q = 2 \times 10^{-8} \mathrm{C}\) on each plate, the capacitor must have a voltage difference \(\Delta V\) such that: $$ Q = C \Delta V. $$ Therefore, we can calculate the voltage difference as: $$ \Delta V = \frac{Q}{C} = \frac{2 \times 10^{-8} \mathrm{C}}{1.03 \times 10^{-10} \mathrm{F}} = 194 \mathrm{V}. $$ Now, the electric field \(E\) is related to the voltage difference and plate separation as follows: $$ E = \frac{\Delta V}{d}. $$ By substituting the given plate separation and the calculated voltage difference, we can find the electric field: $$ E = \frac{194 \mathrm{V}}{1 \times 10^{-3} \mathrm{m}} = 1.94 \times 10^5 \mathrm{V/m}. $$ Now we have the answers to both parts of the exercise: (a) The capacitance of this capacitor is \(1.03 \times 10^{-10} \mathrm{F}\). (b) The electric field that must be applied for \(2 \times 10^{-8} \mathrm{C}\) to be stored on each plate is \(1.94 \times 10^5 \mathrm{V/m}\).