Question

A parallel-plate capacitor has circular plates of 8.20 cmradius and 1.30 mmseparation. (a) Calculate the capacitance. (b) Find the charge for a potential difference of 120 V

Short answer

a) The capacitance is 144 pF.

b) The charge for a potential difference 120 V is 17.3 nC

Step-by-step solution

Given

The parallel-plate capacitor has circular plates of radius,

$\mathrm{R}=8.20\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=8.20\times {10}^{-2}\mathrm{m}$.

The separation between the plates is$\mathrm{d}=1.30\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)=1.30\times {10}^{-3}\mathrm{m}$

The potential difference is V = 120 V.

Determining the concept

By using Eq.25-9 and 25-1, find the capacitance and the charge for the potential difference 120 V.

Formulae are as follows:

$$\begin{gathered} \mathbf{C}\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}} \\ \mathbf{A}\mathbf{=}{\mathbf{\tau \tau R}}^{\mathbf{2}} \\ \mathbf{C}\mathbf{=}\frac{\mathbf{q}}{\mathbf{V}} \end{gathered}$$

Where${\mathbf{\epsilon }}_{\mathbf{0}}$is permittivity, Ais the area of capacitor plates, and dis the separation between the plates, Ris the radius of the circular plate,V is the potential difference, and q is the charge on the particle.

(a) Determining the capacitance

From the equation.25-9, the capacitance of the parallel plate capacitor is given by,

$\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$

Where${\mathrm{\epsilon }}_0$is permittivity, A is the area of capacitor plates, and d is the separation between the plates.

The plates are circular, so, the plate area is given by,

$\mathrm{A}={\mathrm{\tau \tau R}}^2$

Where R is the radius of the circular plate.

Substituting the given values in the above equation,

$$\begin{gathered} \mathrm{C}=\frac{{\mathrm{\epsilon }}_0{\mathrm{\tau \tau R}}^2}{\mathrm{d}} \\ =\frac{\left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right)\times 3.142\times {\left(8.20\times {10}^{-2}\mathrm{m}\right)}^2}{1.30\times {10}^{-3}\mathrm{m}} \\ =1.44\times {10}^{-10}\mathrm{F}\left(\frac{1\mathrm{pF}}{{10}^{-12}\mathrm{F}}\right) \\ =144\mathrm{pF} \end{gathered}$$

Hence, the capacitance is 144 pF.

(b) Determining the charge for a potential difference of

From the equation.25-1, the charge on the positive plate is given by,

$\mathrm{q}=\mathrm{CV}$

Where V is the potential difference.

$$\begin{gathered} \mathrm{q}=\left(1.44\times {10}^{-10}\mathrm{F}\right)\times \left(120\mathrm{V}\right) \\ =1.73\times {10}^{-8}\mathrm{C}\left(\frac{1\mathrm{nC}}{{10}^{-9}\mathrm{C}}\right) \\ =17.3\mathrm{nC} \end{gathered}$$

Hence, The charge for a potential difference 120 V is 17.3 nC.

Therefore, by using the formula of the capacitance of the parallel plate, capacitance and potential difference can be determined.