Polystyrene has dielectric constant 2.6 and dielectric strength 2.0x10^7 V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

Verified step by step guidance

1. Step 1: Calculate the electric field (E) when it is 80% of the dielectric strength. Use the formula E = 0.8 \times \text{dielectric strength}.
2. Step 2: Use the formula for energy density (u) of a capacitor with a dielectric, which is u = \frac{1}{2} \epsilon_0 \kappa E^2, where \epsilon_0 is the permittivity of free space (8.85 \times 10^{-12} F/m) and \kappa is the dielectric constant.
3. Step 3: For part (b), calculate the electric field (E) again using the same method as in Step 1 since the conditions are the same.
4. Step 4: Use the formula for the capacitance of a parallel-plate capacitor with a dielectric, C = \kappa \epsilon_0 \frac{A}{d}, where A is the area of the plates and d is the separation between them. Solve for A using the relationship between capacitance, stored energy (U), and voltage (V): U = \frac{1}{2} C V^2.
5. Step 5: Rearrange the formula from Step 4 to solve for the area (A) of each plate, using the given values for the stored energy, voltage, and calculated capacitance.