Question

The capacitor in Fig. 25-25 has a capacitance of$\mathbf{25}\mathbf{\mu F}$ and is initially uncharged. The battery provides a potential difference of 120 VAfter switchis closed, how much charge will pass through it?

Short answer

Charge passing through the battery is$\mathrm{q}=3.0\times {10}^{-3}\mathrm{C}$

Step-by-step solution

Given

The capacitance of the capacitor is$\mathrm{C}=25\mathrm{\mu F}\left(\frac{{10}^{-6}\mathrm{F}}{1\mathrm{\mu F}}\right)=2.5\times {10}^5\mathrm{F}$.

The potential difference is V = 120 V.

Determining the concept

Using equation 25-1, find the amount of charge that passes through the battery.

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.

(a) Determining the charge passing through the battery

From equation 25-1 we know that,

q = CV

The charge passing through the battery is,

$$\begin{gathered} \mathrm{q}=\left(2.5\times {10}^{-5}\mathrm{F}\right)\left(120\mathrm{V}\right) \\ =3.0\times {10}^{-3}\mathrm{C} \end{gathered}$$

Hence, the charge passing through the battery is$3.0\times {10}^{-3}\mathrm{C}$

Therefore, by using the formula for capacitance find the charge passing through the battery.