Question

The two circuits shown in Figure 19.59 have different capacitors but the same batteries and thin-filament bulbs. The capacitors in circuit $\mathbf{1}$and circuit $\mathbf{2}$areidentical exceptthat the capacitor in circuit $\mathbf{2}$was constructed with its plates closer together. Both capacitors have air between their plates. The capacitors are initially uncharged. In each circuit the batteries are connected for a short time compared to the time required to reach equilibrium, and then they are disconnected. In which circuit (1or 2) does the capacitor now have more charge? Explain your reasoning in detail.

Short answer

The Capacitor$\mathbf{2}$ has more charge.

Step-by-step solution

Write the given data from the question.

The two circuits consist batteries, thin filaments bulbs and Capacitors.

Distance between the plates of the Capacitor $\mathbf{2}$is less as compare to the capacitor $\mathbf{1}$.

Determine the formulas to calculate the capacitor have more charge.

The expression to calculate the capacitance of the capacitor is given as follows.

$\mathit{C}\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}$

Here, $\mathit{A}$is the area of the plates, and $\mathit{d}$is the separation between the plates.

The expression to calculate the charge on the plates is given as follows.

$\mathit{Q}\mathbf{=}\mathit{C}\mathit{\Delta }\mathit{V}$ …… (i)

Here,$\mathit{\Delta }\mathit{V}$is the potential difference between the plates of capacitor.

Calculate the capacitor have more charge.

Calculate the charge on the plates of the capacitors.

Substitute$\frac{{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}$for$\mathit{C}$into equation (i).

$\mathit{Q}\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}\mathit{\Delta }\mathit{V}$

Since all the parameters of the circuit are the same therefore, the term ${\mathit{\epsilon }}_{\mathbf{0}}\mathit{A}\mathit{\Delta }\mathit{V}$can be assumed as constant and equation can be rewrite as,

$\mathit{Q}\mathbf{}\mathit{\alpha }\frac{\mathbf{1}}{\mathbf{d}}$

From the above expression, it is clear that the charge on the capacitor plates is inversely proportional to the distance between the plates of the capacitor.

Therefore, the capacitor with less distance between the plates will have more charge. Capacitor$\mathbf{2}$ has less distance and will have more charge than capacitor$\mathbf{1}$ .

Hence capacitor$\mathbf{2}$ has more charge.