A5.00-pf parallel-plate air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00×102v. the electric field between the plates is to be no greater than 1.00×104n/c. as a budding electrical engineer for live-wire electronics, your task is to design the capacitor by finding what its physical dimensions and separation must be.

A 5 pF parallel-plate air-filled capacitor, with a potential difference equal to 100 Volts and an electric field of 10.000 Volts per meter, must have a plate separation of 10 milimeters, and plate's area equal to 0.0056 meters squared.

Capacitors are electrical components use to store electrical energy in the form of potential difference. This is done by putting 2 plates of a conductive material very close to each other, but without getting them into contact. The idea is that, when you charge one plate with (for example) positive charges, the other plate will "feel" this and will charge itself with negative charges (due to the attraction between charges of opposite signs).

The trick is that, even though the charges in the two plates attract each other, they can't come into contact since the 2 plates are separated by a certain distance. Now let's suppose there is only air between the capacitor's plates, we know that air is not a conductive material, however if you charge both plates just enough, you'll be able to make air conduct electricity and so charges from one plate will go to the other and equilibrium will be achieved.

This is the same thing that happens when a lightning strikes the ground. Imagine that the Earth is a gigantic capacitor, with the clouds being one plate and the ground being the other plate. When the clouds are charged enough, air is forced to conduct electricity towards the ground (which is the lightning we see), until equilibrium is achieved. This is the reason why, in Physics, we say that everything is a conductive material if you force it just enough.

Back to our problem... We can compute the separation of the capacitor's plates with the following equation:

Which says that the Voltage V is equal to the product of the Electric field E and the distance between the plates d. This equation is valid if the electric field across the capacitor is constant (which for small magnitudes of d it is). So solving for the distance we get:

Where we can see that the separation of plates is 10 millimeters. Now for the area between the plates we can use this other formula:

Where is the air's permittivity which has a value of around . By solving for the plate's area, we can find:

Which corresponds to a plate's radius of 42.4 millimeters.

Capacitor, Electric charge, Electric field, Voltage

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

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explanation:

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