High School

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A parallel-plate capacitor has circular plates with a radius of 7.54 cm and a separation of 1.71 mm.

(a) Calculate the capacitance.

(b) What charge will appear on the plates if a potential difference of 97.7 V is applied?

Answer :

To calculate the capacitance of a parallel-plate capacitor and the charge on its plates when a potential difference is applied, we'll follow these steps:

(a) Calculate the capacitance.

The capacitance [tex]C[/tex] of a parallel-plate capacitor is given by the formula:

[tex]C = \frac{\varepsilon_0 \cdot A}{d}[/tex]

Where:


  • [tex]\varepsilon_0[/tex] is the vacuum permittivity, approximately [tex]8.85 \times 10^{-12} \ \text{F/m}[/tex].

  • [tex]A[/tex] is the area of one of the plates.

  • [tex]d[/tex] is the separation between the plates.



  1. Calculate the area [tex]A[/tex]: The plates are circular, so the area [tex]A[/tex] is:

    [tex]A = \pi r^2[/tex]

    With [tex]r = 7.54 \ \text{cm} = 0.0754 \ \text{m}[/tex]:

    [tex]A = \pi (0.0754)^2
    \approx 0.0179 \ \text{m}^2[/tex]


  2. Plug the known values into the formula:

    The separation [tex]d = 1.71 \ \text{mm} = 0.00171 \ \text{m}[/tex].

    [tex]C = \frac{(8.85 \times 10^{-12} \ \text{F/m}) \times 0.0179 \ \text{m}^2}{0.00171 \ \text{m}}
    \approx 9.25 \times 10^{-12} \ \text{F} \text{ or } 9.25 \ \text{pF}[/tex]



(b) What charge will appear on the plates if a potential difference of 97.7 V is applied?

The charge [tex]Q[/tex] on the plates is related to the capacitance and the potential difference [tex]V[/tex] by the formula:

[tex]Q = C \cdot V[/tex]

Using the capacitance we calculated:

[tex]Q = (9.25 \times 10^{-12} \ \text{F}) \times 97.7 \ \text{V}
\approx 9.03 \times 10^{-10} \ \text{C} \text{ or } 903 \ \text{pC}[/tex]

So, the capacitance of the capacitor is approximately [tex]9.25 \ \text{pF}[/tex], and the charge on the plates when a [tex]97.7 \ \text{V}[/tex] potential difference is applied is approximately [tex]903 \ \text{pC}[/tex].

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Rewritten by : Jeany