How Does Changing Plate Radius Affect Capacitance in a Parallel Plate Capacitor?

AI Thread Summary
The discussion focuses on how the radius of plates in a parallel plate capacitor affects capacitance. To achieve a capacitance of 2.8 microfarads with a plate separation of 1.6mm, the required radius can be calculated using the formula C = ε*A/d, where A is the area of the plates. If the separation between the plates decreases, the radius should be increased to maintain the same capacitance, as capacitance is directly proportional to the area of the plates. For a plate separation of 3mm, the radius can also be determined using the same capacitance formula. Understanding the relationship between permittivity, area, and distance is crucial for solving these problems.
papi
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Consider a parallel plate capacitor constructed from two circular metal plates of radius R. The plates are separated by a distance of 1.6mm

a. What radius must the plates have if the capacitence is 2.8 micro F?

b. if the separation between the plates is decreased, should the radius of the plates be increased or decreased to maintain the capacitence? PLEASE EXPLAIN

c. Find the radius of the plates that gives a capacitence of 2.8 micro F for a plate separation of 3 mm

I know C=Q/V but how does that help here?
 
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i still don't get it at all. would d=2R and I don't know the E here.. it isn't given
 
papi said:
i still don't get it at all. would d=2R and I don't know the E here.. it isn't given

Throw me a bone here. It's not E. It's ε - permittivity. It would be εo for air. A is area (the radius might help with that) and d is the distance between the plates.

Try reading the link again if you are confused.
 

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