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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 81.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00×106 V/m.
What is the potential difference between the plates?
What is the area of each plate in cm2?
What is the capacitance of the capacitor in picofarads?

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cjrodriguez89

Answer:

a) V = 10⁴ V = 10000 V

b) A =22.9 cm²

c) C = 8.1 pF

Explanation:

a)

  • The potential difference V between the plates, is related with the electric field E, produced by the charge on one of the plates, as follows:

        [tex]V = E*d[/tex]

  • where d = distance between plates= 2.5*10⁻³ m.
  • Replacing by the givens of E and d, we can find V as follows:

        [tex]V = 4.00e6 V / m* 2.5e-3 m = 10e4 V = 10000 V[/tex]

b)

  • Applying Gauss 'law to a gaussian surface like a pillbox half outside one of the plates, one half inside it, as the electric field must be normal to the surface, and constant due to the surface charge density is uniform, we can write the following equation:

        [tex]E*A = \frac{Q}{\epsilon_{0} }[/tex]

  • Solving for A:

       [tex]A = \frac{Q}{E*\epsilon_{0}} = \frac{81e-9C}{4.00e6N/C*8.85e-12(C2/N*m2)}}\\ A= 22.9 cm2[/tex]

c)

  • By definition, the capacitance of a capacitor, is given by the following equation:

       [tex]C = \frac{Q}{V}[/tex]

  • Replacing Q= 81*10⁻⁹C, and V = 10⁴V, we have:

       [tex]C = \frac{Q}{V} =\frac{81e-9C}{10e4 V} = 8.1e-12 F[/tex]

  • The capacitance of the capacitor is 8.1 pF.

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