A parallel-plate capacitor has capacitance C0 = 5.00 pF

Solution 36E Step 1 of 5: The given parallel plate capacitor has capacitance C = 0 pF when it has air between plates and separation between the plates is d= 1.5 mm. We need to find the Part(a) maximum magnitude of the charge on each plate Q = , if the maximum applied field max between the plates is E = 3 × 10 V /m. In the part (b) When a dielectric with K = max 2.70 is inserted between the plates of the capacitor, we need to calculate the maximum magnitude of the charge on each plate Q 1 = , if the maximum applied field between max the plates is E = 3 × 10 V /m. max Given data, Capacitance of the capacitor(air), C = 5 p F = 5 × 1012F 0 Distance between the plates, d= 1.5 mm=1.5 × 10 m3 Dielectric constant, K = 2.70 Applied maximum electric field, E = 3 × 10 V /m max To find, (a) (Between plates air)maximum magnitude of the charge on each plate Q max = (b) (Between plates dielectric)maximum magnitude of the charge on each plate Qmax 1 =