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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 :

cjrodriguez89

Answer:

6.48*10⁻⁷ C

Explanation:

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

       [tex]C =\frac{Q}{V}[/tex]

  • 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:

        [tex]V = E_{max} * d[/tex]

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

        [tex]V = 6.0e6 V/m * 0.00108 m= 6480 V[/tex]

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

        [tex]Q_{max} = C*V = 1e-10F* 6480 V = 6.48e-7 C[/tex]

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

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

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