Question

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?

Answer 1

Answer:

7.2 x 10⁻⁷C

Explanation:

From Gauss's law, the maximum charge Q, of a capacitor is given by

Q =  E x A x ε₀       -----------------------(i)

Where;

A = surface area = 4 x π x r                [r is thickness or radius]

ε₀ = permittivity of free space or air = 8.85 x 10⁻¹²F/m

E = electric field

From the question;

r = 1.08mm = 0.00108m          [Take π = 3.142]

A  = 4 x π x 0.00108 = 0.01357m²

E = 6.0 x 10⁶V/m

Substitute these values into equation (i) as follows;

Q = 6.0 x 10⁶ x 0.01357 x 8.85 x 10⁻¹²

Q = 7.2 x 10⁻⁷C

Therefore, the maximum charge it can hold is 7.2 x 10⁻⁷C

Answer 2

Answer:

6.48*10⁻⁷ C

Explanation:

• By definition, the capacitance of a capacitor is expressed as follows:

       $C =\frac{Q}{V}$

• where Q is the charge on one of the plates of the capacitor, and V the potential difference between the plates.
• The maximum electric field, the potential difference, and the distance between plates are related by the following expression:

        $V = E_{max} * d$

• Replacing by the givens, we can find V as follows:

        $V = 6.0e6 V/m * 0.00108 m= 6480 V$

• Now, we can find the maximum charge Qmax,  as follows:

        $Q_{max} = C*V = 1e-10F* 6480 V = 6.48e-7 C$

• The maximum charge that the capacitor can hold is 6.48*10⁻⁷ C.