# What is the initial capacitance of a computer keyboard key?

• stylez03
In summary: C_0 = \frac {e_0 A} {s_0} = \frac {(8.85 \times 10^{-12} F/m)(50.5 \times 10^{-6} m^2)}{(0.660 \times 10^{-3} m)} = 0.676 pFThen if \Delta C = 0.300 pF, you can plug that into your equation and solve for \Delta s. You've got the right idea, just need to clean up some notation and be careful with your units.In summary, in a computer keyboard with a parallel-plate capacitor system, the change in capacitance when a key is depressed can be detected by electronic

## Homework Statement

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, air-filled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 50.5 mm^2, and the separation between the plates is 0.660 mm before the key is depressed.

## Homework Equations

$$\Delta C = \frac {e_{0}A} {s_{1}} - \frac {e_{0}A} {s_{0}}$$

## The Attempt at a Solution

$$\Delta s = s_{0} - s_{1} => s_{0} *(1 - \frac {1} {1 + \frac {s_{0} \nabla C} {e_{0}A}})$$

I'm not too sure if this is correct, and I'm very bad at the unit conversion and I think that also might be an issue

is $$C = 0.300 * 10^{-12}$$

Last edited:
Sorry, you haven't stated the full problem. What are you supposed to do? Are you given a final (smaller) separation distance, and asked to find the initial and final capacitances? Or given a target change in capacitance, and asked to solve for the final separation distance?

Also, the "Delta" symbol would usually be used for "change". You've used the "del" or "nabla" symbol above in your post, and del has a different meaning than Delta. Small nitpick, but it confused me when I first saw your post.

Delta = $$\Delta$$

Finally, to help you with units and unit conversions, just multiply them out like they were variables. Like, velocity is distance per time, or:

$$v [m/s] = \frac{\Delta x [m]}{\Delta t }$$

And if you need to convert microseconds to seconds or something, just multiply through by one (like 1s = 10^6us), like this:

$$\Delta t = \Delta t [\mu s] \frac{1 }{10^6 [\mu s]}$$

Sorry I left out the actual question:

If the circuitry can detect a change in capacitance of 0.300 pF, how far must the key be depressed before the circuitry detects its depression?

So what do you get for the initial capacitance?

## What is capacitance?

Capacitance is a measure of the ability of a system to store an electric charge. It is defined as the ratio of the amount of charge stored on a conductor to the potential difference between the conductors.

## How does capacitance change with distance?

Capacitance is inversely proportional to the distance between the conductors. This means that as the distance between two conductors decreases, the capacitance increases, and vice versa.

## What factors affect the capacitance of a system?

The capacitance of a system is affected by the area of the conductors, the distance between them, and the material between the conductors. It also depends on the type of material used for the conductors and the dielectric constant of the medium between them.

## What is the formula for calculating capacitance?

The formula for calculating capacitance is C = Q/V, where C is the capacitance in farads, Q is the charge in coulombs, and V is the potential difference in volts.

## Can capacitance be changed?

Yes, capacitance can be changed by altering the physical parameters of the system, such as the distance between the conductors or the area of the conductors. It can also be changed by changing the dielectric material between the conductors.

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