Chapter 24, Problem Q24.1DQ

Equation (24.2) shows that the capacitance of a parallel-plate capacitor becomes larger as the plate separation d decreases. However, there is a practical limit to how small d can be made, which places limits on how large C can be. Explain what sets the limit on d. (Hint: What happens to the magnitude of the electric field as d → 0?)

To determine

The parameter which sets the limit of $d$.

Explanation of Solution

When the distance existing between plates increases, the electric field will be beyond the limit of dielectric strength of the material between the plates. In this case, the current flow through the material and the capacitor gets discharge.

Formula to calculate the electric field on the plates is,

$E=\frac{V}{d}$ (I)

• $V$ is the potential difference,
• $d$ is the distance existing between the plates.

From equation (I), the electric field and distance existing between plates are inversely proportional. When the distance is very small the electric field will be beyond the limit of dielectric strength of the material between the plates. In this case, the current flow through the material and the capacitor gets discharge.

Therefore, the electric field sets the limit of $d$.

Conclusion:

Thus, the parameter which sets the limit of $d$ is electric field.