Question

Consider two capacitors whose only difference is that the plates of capacitor number 2 are closer together than those of capacitor number 1 (Figure 19.56). Neither, capacitors has an insulating layer between the plates. They are placed in two different circuits having similar batteries and bulbs in series with the capacitor.

Show that in the first fraction of a second, the current stays nearly constant (decreases less rapidly) in the circuit with capacitor number 2. Explain your reasoning in detail.

Hint: Show charges on metal plates, and consider the electric fields they produce in the nearby wires. Remember that the fringe field near a plate outside a circular capacitor is approximately-

$\mathbf{\left(}\frac{{\displaystyle \raisebox{1ex}{$Q$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\mathbf{\epsilon }}_{\mathbf{o}}}\mathbf{\right)}\left(\frac{s}{2R}\right)$

More extensive analysis shows that this trend holds true for the entire charging process: the capacitor with the narrower gap ends up with more charge on the plates.

Short answer

It is shown that if the separation between the plates of a capacitor is less, the current will vary less rapidly.

Step-by-step solution

Given Data

The plates of capacitor $2$ are closer to each other as compared to capacitor$1$.

Capacitance

The ability of a capacitor to store energy in the form of an electric field is known as capacitance. It is the ratio of charge accumulated on the plate to the potential difference developed between the plates.

Calculation

Consider a parallel plate capacitor, having an area of cross-section ‘A’. The separation between the plates is s, and there is no layer of insulating material between them. Thus, the material between the plates is free space. When connected to a circuit, the electric field, developed between the plates, is given as-

$$\begin{gathered} E=\frac{\sigma }{{\epsilon }_o} \\ =\frac{{\displaystyle \raisebox{1ex}{$Q$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\epsilon }_o} \end{gathered}$$

Here, $Q$is the charge gathered on the plates,localid="1662168754855" ${\epsilon }_o$is the permittivity of free space and localid="1662168738233" $\sigma$is the surface charge density.

The potential difference$\left(∆V\right)$developed between the plates is given as-

localid="1662170000699" $$\begin{gathered} ∆V=E.s \\ =\left(\frac{{\displaystyle \raisebox{1ex}{$Q$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\epsilon }_o}\right)s \end{gathered}$$

In the above equation, the potential difference produced between the plates is directly proportional to the separation between the plates. The larger the separation, the larger will be the potential difference, and the higher will be the rate of accumulation of charge. Hence, the current will decrease more rapidly.

Therefore, if the separation between the plates of the parallel plate capacitor is less, the current will decrease less rapidly.