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 1.16-mm thickness of Rutile can hold? A) 1.8 nC B) 0.7 µC C) 0.3 mC D) 6.96 C E) 60 nC

Question

Quizplus · Accepted Answer

The maximum charge a capacitor can hold is given by:

Q = εA(V/d)

Where Q is the charge, ε is the permittivity of the material (in this case, Rutile), A is the area of the plates, V is the voltage across the plates, and d is the distance between the plates.

We are given that the capacitance is 10-11 F and the thickness of Rutile is 1.16 mm = 1.16 x 10^-3 m.

Since capacitance is given by:

C = εA/d

We can rearrange this equation to solve for A:

A = Cd/ε

Substituting the given values, we have:

A = (10^-11 F)(1.16 x 10^-3 m)/(6.0 x 10^9 V/m) = 1.95 x 10^-20 m^2

The maximum charge the capacitor can hold is then:

Q = εA(V/d) = (6.0 x 10^9 V/m)(1.95 x 10^-20 m^2)(6.0 x 10^7 V/m) = 0.7 µC

Therefore, the answer is B.

The Dielectric Strength of Rutile Is 6