Equation (24.2) shows that the capacitance of a parallel plate capacitor becomes larger as the plate separation ddecreases. However, there is a practical limit to how small dcan be made, which places limits on how large Ccan be. Explain what sets the limit on d. (Hint:What happens to the magnitude of the electric field as d0?)

Capacitance is the measure or the capacity of an electric device to store charge and allow the device to use the charge up as desired. The difference in potential causes the charge to get stored on the plates of conductor.

Mathematically,

$\mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{V}}$

Where, Q is the total charge stored and V is the potential difference or the voltage applied electronically.

According to the given equation 24.2, capacitance is inversely proportional to the distance d, i.e, as d decreases the capacitative action increases.

Now when $d\to 0$, then the equation becomes,

$$\begin{gathered} \mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{0} \\ =\infty \end{gathered}$$

In practical scenario such high capacitance gives rise to short circuiting with no potential difference and so there has to be a limit to the lowering of d value so that the plates don’t touch. The electric field has a similar reaction when d is made zero and it blows up ionizing the surrounding air and causing dielectric breakdown. Therefore, the distance between the plates is limited.