A 1-megabit computer memory chip contains many 60.0 -fF...

A 1-megabit computer memory chip contains many $60.0$ -fF capacitors. Each capacitor has a plate area of $21.0 \times 10^{-12} \mathrm{~m}^{2}$. Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The characteristic atomic diameter is $10^{-10} \mathrm{~m}=$ $0.100 \mathrm{~nm}$. Express the plate separation in nanometers.

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by an insulating material (dielectric). Its capacitance is given by the formula C = ?A/d, where C is the capacitance, ? is the permittivity of the medium between the plates, A is the area of one plate, and d is the separation distance. This relationship shows how the physical configuration directly affects the capacitor's ability to store electrical charge.

Permittivity of Free Space

The permittivity of free space, denoted as ??, is a fundamental physical constant that quantifies the ability of a vacuum to permit electric field lines. It is used in the calculation of capacitance for devices operating in air or vacuum, acting as a scaling factor that relates geometry (area and separation) to capacitance.

Capacitance Formula and Geometric Relationship

The formula C = ?A/d encapsulates the direct relationship between the capacitance of a device and its geometric dimensions. It implies that increasing the plate area or decreasing the separation distance will result in a higher capacitance, underscoring important design considerations in electronic component fabrication.

Unit Conversion and Dimensional Analysis

Ensuring that all quantities are expressed in consistent units is crucial in physics calculations. This includes converting units of capacitance (like femtofarads to farads) and area (to square meters) so that the formula components align dimensionally, allowing for an accurate calculation of physical quantities such as the separation distance.

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