need help on how to solve 67. The dielectric strength of Rutile is 6.0 times 10^

Question

need help on how to solve 67.

The dielectric strength of Rutile is 6.0 times 10^6 V/m, which corresponds to the maximum electric field at the dielectric can sustain before breakdown. What is the maximum charge that a 10^-10-F capacitor with a 1.0-mm thickness of Rutile can hold? 1.7 nC 0.60 muC 0.30 mC 6.0 C A parallel-plate capacitor has dimensions 4.0 cm times 5.0 cm. The plates are separated by a 1.0-mm thickness of paper (dielectric constant kappa = 3.7). What is the charge that can be stored on this capacitor, when connected to a 1.5-V battery? (epsilon_0 = 8.85 times 10^-12C^2/N Middle Dot m^2) 20 times 10^-12 C 4.8 times 10^-12 C 4.8 times 10^-11 C 9.8 times 10^-11 C

Explanation / Answer

E =6*10^6 V/m , C = 10^-10 F, d = 1mm

C = Akeo/d

Akeo = Cd

E = Q/Akeo = Q/Cd

Q = 6*10^6*10^-10*1*10^-3

Q = 0.6 uC

Correct option is (b)

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