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 1.08-mm thickness of rutile can hold? A)1.7 nC B)0.65 µC C)0.32 mC D)6.5 C E)60 nC

Question

Quizplus · Accepted Answer

The maximum charge that a capacitor can hold is directly proportional to the dielectric strength of the material and the area of the plates and inversely proportional to the distance between the plates.

Using the formula for capacitance, C = εA/d, where ε is the permittivity of the material, A is the area of the plates, and d is the distance between the plates, we can rearrange to solve for the maximum charge, Q:

Q = V C = V ε A/d

where V is the breakdown voltage, which is equal to the product of the dielectric strength and the distance between the plates:

V = Ed

Substituting the values given in the problem, we have:

V = (6.0 × 10^7 V/m)(1.08 × 10^-3 m) = 6.48 × 10^4 V

And:

Q = (6.48 × 10^4 V)(8.85 × 10^-12 F/m)(π(0.01 m)^2/1.08 × 10^-3 m) = 6.52 × 10^-7 C

Converting to microcoulombs, we get:

Q = 0.65 µC

Therefore, the correct answer is B.

The Dielectric Strength of Rutile Is 6