91Ó°ÊÓ

The capacitance is$\mathrm{C}=25.0\mathrm{μ¹ó}\left(\frac{{10}^{-6}\mathrm{F}}{1\mathrm{F}}\right)=2.50×{10}^{-5}\mathrm{F}$

The potential difference is$\mathrm{V}=4200\mathrm{V}$

Using Eq.25-1, and 25-19, find the amount of charge that passes through meter A.

Formulae are as follows:

$\mathbf{C}\mathbf{=}\frac{\mathbf{q}}{\mathbf{v}}$

Where C is capacitance, V is the potential difference, and q is the charge on the particle.

From the equation 25-1, the charge that passes through meter A is given by, $\mathrm{q}={\mathrm{C}}_{\mathrm{eq}}\mathrm{V}$

Where${\mathrm{C}}_{\mathrm{eq}}$is the equivalent capacitance of parallel arrangement of capacitors with the capacitance of each capacitor as C.,

From the equation 25-19, if capacitors are in parallel, the equivalent capacitance is,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3 \\ =3\mathrm{C} \end{gathered}$$

Therefore,

$$\begin{gathered} \mathrm{q}=3\mathrm{CV} \\ =3×\left(2.50×{10}^{-5}\mathrm{F}\right)×\left(4200\mathrm{V}\right) \\ =0.315\mathrm{C} \end{gathered}$$

Hence, the amount of charge that passes through meter A is q = 0.315 C .

Therefore, by using the equivalent capacitor formula, the charge passing through the meter can be determined.