The dielectric strength of rutile is 6.0 × 10⁶ V/m,which corresponds to the maximum electric field that the dielectric can sustain before breakdown.What is the maximum charge that a 10−¹⁰-F capacitor with a 0.50-mm thickness of Rutile can hold? A) 0.30 μC B) 6.0 C C) 0.60 µC D) 1.7 nC

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Quizplus · Accepted Answer

The maximum charge that a capacitor can hold is given by Q = CV, where C is the capacitance and V is the voltage. The voltage across the capacitor is equal to the electric field times the thickness of the dielectric, so V = Ed. Substituting this into the equation for Q, we get Q = CEd. The capacitance of a parallel-plate capacitor is given by C = εA/d, where ε is the permittivity of the dielectric, A is the area of the plates, and d is the distance between the plates. Substituting this into the equation for Q, we get Q = εAdE. Plugging in the given values, we get Q = (8.85 × 10^-12 F/m)(10^-10 F)(0.50 × 10^-3 m)(6.0 × 10^6 V/m) = 0.30 μC.

The Dielectric Strength of Rutile Is 6