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?

Question

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The maximum charge that a capacitor can hold is given by Q = CV, where C is the capacitance and V is the voltage. 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. The dielectric strength of a material is the maximum electric field that the material can sustain before breakdown. The electric field in a capacitor is given by E = V/d. Therefore, the maximum charge that a capacitor can hold is given by:Q = CV = εAV/d = (εA/d)V = CEwhere E is the dielectric strength of the material.In this problem, the dielectric strength of rutile is 6.0 × 10^6 V/m, the thickness of the dielectric is 0.50 mm = 5.0 × 10^-4 m, and the area of the plates is not given. Therefore, we cannot calculate the maximum charge that the capacitor can hold without knowing the area of the plates. However, we can say that the maximum charge that the capacitor can hold is proportional to the area of the plates.

The Dielectric Strength of Rutile Is 6 μ\muμ C B) 6

The Dielectric Strength of Rutile Is 6 $\mu$ C
B) 6