A capacitor of unknown capacitance Cis charged toand connected across an initially uncharged$\mathbf{60}\mathbf{}\mathbf{\mu F}$capacitor. If the final potential difference across the$\mathbf{60}\mathbf{}\mathbf{\mu F}$capacitor is 40 V, what is C?

The capacitor which is connected parallel with C when C is fully charged.

$$\begin{gathered} a)C\text{'}=60\times {10}^{-6}F \\ b)V=100V \end{gathered}$$

The law of conservation of the charge states that the total charges are conserved, we cannot create or destroy the charges. The charge on the capacitor is equal to the product of capacitance and voltage of the capacitor.

We will use conservation of charge, in this case, to find the capacitance.

Formulae:

The charge stored between the plates of the capacitor, q = CV …(i)

According to the conservation of charge,

$Q=q_1+q_2$ (2)

Let Q be the charge on the capacitor C when it is fully charged to potential V = 100 V

This equation implies the following condition derived from the formula of equations (i) and (ii) as:

$$\begin{gathered} \mathrm{C}\times 100\mathrm{V}=\mathrm{C}\times 40\mathrm{V}+\mathrm{C}\text{'}\times 40\mathrm{V} \\ \mathrm{C}\times (100-40)=\mathrm{C}\text{'}\times 40 \\ \mathrm{C}\times 60=\times 60\times {10}^{-6}\mathrm{F}\times 40 \\ \mathrm{C}=\frac{60\times {10}^{-6}\mathrm{F}\times 40}{60} \\ =40\times {10}^{-6}\mathrm{F}\mathrm{or}40\mathrm{\mu F} \end{gathered}$$

Hence, the value of the capacitance is $40\mathrm{\mu F}$.