Question

3. The paper dielectric in a paper and aluminum foil capacitor is 0.005 cm thick. It's dielectric coefficient is 2.5 and its dielectric strength is 5.0 x 107 V/m. What area of paper and foil is required for a 0.1 HF capacitor? If the electric field between the plates is not to exceed one half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?

Answer 1

Capacitance, C = K A E_0/d A: area of paper and fail d: thickness = 0.005 cm = 0.005 times 10^- 2 m k: dielectric constant = 2.5 0.1 times 10^- 6 = (2.5 A) (8.854 times 1^- 12 F/M)/0.005 times 10^- 2 A = 0.22588 m^2 Dielectric strength rightarrow E = 5 times 1^7v/m potential difference = (E/2) d Delta V = (5 times 10^7 v/m/2) (0.005 times 10^- 2 m) Delta v = 1250 volts