A 100 pFcapacitor is charged to a potential difference of 50 V,and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential difference across the first capacitor drops to 35 Vwhat is the capacitance of this second capacitor?

The capacitance ${\mathrm{C}}_1=100\mathrm{pF}$.

The potential difference${\mathrm{V}}_0=50\mathrm{V}$

Capacitors 1 and 2 are connected parallel by removing the charging battery.

The capacitor 1 drop to the potential difference V = 35 V.

Using the equation 25-1 and applying the conservation of charge, find thecapacitance of the second capacitor.

The law of conservation of charge states that electric charge can neither be created nor destroyed in a closed system, but the amount of charge remains the same.

Formulae are as follows:

From the equation 25-1, the charge on the charged capacitor is given by, q = CV

The charge conservation gives,${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{q}\mathbf{-}{\mathbf{q}}_{\mathbf{1}}$

The capacitance${\mathbf{C}}_{\mathbf{2}}$ is,${\mathbf{C}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{2}}}{\mathbf{V}}\mathbf{a}$

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

From the equation 25-1, the initial charge on the capacitor is given by,

$\mathrm{q}={\mathrm{C}}_1{\mathrm{V}}_0$

where${\mathrm{C}}_1=100\mathrm{pF}$ and V = 50 V.

After the battery is disconnected and the second capacitor is wired in parallel to the first, the charge on the first capacitor is,

${\mathrm{q}}_1={\mathrm{C}}_1\hspace{0.17em}\mathrm{V}$

Where, V = 35 V

Since the charge is conserved in this process, the charge on the second capacitor is,

${\mathrm{q}}_2=\mathrm{q}-{\mathrm{q}}_1$

Where,$C_2$is the capacitance of the second capacitor.

Substituting,and, giving,

$$\begin{gathered} q_2=C_1{V}_0-C_{1}V \\ =C_1\left(V_0-V\right) \end{gathered}$$

The potential difference across the second capacitor is also. Therefore, the capacitanceis,

${\mathrm{C}}_2=\frac{{\mathrm{q}}_2}{\mathrm{V}}$

${\mathrm{C}}_2=\frac{{\mathrm{C}}_1\left({\mathrm{V}}_0-\mathrm{V}\right)}{\mathrm{V}}$

Therefore,

$$\begin{gathered} {\mathrm{C}}_2=\frac{\left(100\mathrm{pF}\right)\left(50\mathrm{V}-35\mathrm{V}\right)}{35\mathrm{V}} \\ =43\mathrm{pF} \end{gathered}$$

Hence, the capacitance of the second capacitor is ${\mathrm{C}}_2=43\mathrm{pF}$.