So the capacitance is found from Equation 17-7, with the voltage given by 17-4. So Q here is equal to see Time's V, which is equal to see Time's the electric field times the distance. D Therefore, the capacitance, uh, go ahead and write that down here C is equal to Q divided by the electric field. E times the distance D, Which queue here is equal to 0.6 75 times 10 to the minus six. Cool ums. The electric field e is equal to 8.24 times 10 to the fourth. And that's gonna be volts per meter that's gonna be multiplied by the distance D, which is equal to 1.95 times 10 to the minus three meters. Oh, so plugging those values in this expression, we find out the capacitance C is equal to 4.2 times 10 to the minus nine fair acts so we can box it in this part of our solution to the question. Okay, so now the play area here is gonna be found from Equation 17-9, which says that the past in C is equal to the dialect. Her constant K of the material times? Absolutely not. The vacuum permeated the of free space times, the area divided by the distance between D. So the area A is equal to the capacitance See which we've just found times D same distance in the previous part when we solved for the capacitance divided by the dye electric constant K, which were told, is equal to 3.75 times Absalon not which is the Vacuum Perm utility of free space. It's equal to 8.85 times 10 to the minus 12 Coolum squared times Newton per meter squared and this is equal to 0.247 This is gonna be meter squared weaken box. That in is the solution.