Question

The circuit shown in Figure 19.61 consists of two flashlight batteries, a large air-gap capacitor, and Nichrome wire. The circuit is allowed to run long enough that the capacitor is fully charged with $\mathbf{+}\mathbf{}\mathit{Q}$and$\mathbf{-}\mathit{Q}$ on the plates.

Next you push the two plates closer together (but the plates don’t touch each other). Describe what happens, and explain why in terms of the fundamental concepts of charge and field. Include diagrams showing charge and field at several times.

Short answer

The decrease in the spacing between the plates increases the capacitance, electric field and charge of the capacitor.

Step-by-step solution

Write the given data from the question.

The capacitor is fully charged with$\mathbf{+}\mathbf{}\mathit{Q}$ and$\mathbf{-}\mathbf{}\mathit{Q}$ on the plates.

The capacitor plates push together but they don’t touch each other.

Determine the formula to effect on the concept of charge and field.

The electric field is defined as the ratio of the charge and the product of the area of the plates and permeability of the space.

The expression to calculate the electric field is given as follows.

$\mathit{E}\mathbf{=}\frac{\mathbf{Q}}{{\mathbf{A\epsilon }}_{\mathbf{0}}}$ …… (i)

Here,$\mathit{Q}$ is the charge of the capacitor,$\mathit{A}$ is the area of the plates and${\mathit{\epsilon }}_{\mathbf{0}}$ is the permeability of the space.

The expression to calculate the potential difference of the plates is given as follows.

$\mathit{\Delta }\mathit{V}\mathbf{=}\mathit{E}\mathit{d}$ …… (ii)

Here, $\mathit{E}$ is the electric field and$\mathit{d}$ is the spacing between the plates.

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

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

Determine the effect on the concept of charge and field.

Calculate the electric field between the plates.

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

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

Substitute $\mathit{C}$for$\mathit{Q}\mathbf{/}\mathbf{∆}\mathit{V}$into above equation.

$\mathit{C}\mathbf{=}\frac{\mathbf{A}{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{s}}$ …… (iv)

From equation (iv), it is clear that the spacing between the capacitor plates is inversely proportional to the capacitance and capacitance is directly proportional to the charge on the capacitor (from equation (iii)), and the charge is proportional to the electric field (from equation (i)). Therefore, capacitance is proportional to the electric field and charge. A decrease in the spacing between the plates increases the capacitance, electric field, and charge of the capacitor.

Consider the figure shown below describe the effect of changing the spacing between the plates.

Hence the decrease in the spacing between the plates increases the capacitance, electric field and charge of the capacitor.