Question

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?

Short answer

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

Step-by-step solution

Step 1: Given

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.

Determining the concept

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.

Determining the capacitance of the second 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 ${\mathrm{C}}_2=43\mathrm{pF}$.