Physics - Parallel Plate Capacitors help?

Given a plate's area and distance between each other. It is also connected to a 'x'-Volt battery.

How does one go about getting its capacitance? the charges on each plate?

Update:

so, in that case, no matter what voltage i connect it to, it doesn't chance capacitance?

2 Answers

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  • 9 years ago
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    For the first part of your question, if your capacitor has air or a vacuum between

    the plates, the capacitance C = A times Eo divided by d, where A is the plate area,

    d is the distance between them, and Eo is the electric permittivity of free space,

    which is 8.85418 * 10**-12 farads per meter, or 8.85418 picofarads per meter.

    If you have an insulator or dielectric between them, you must include its dielectric

    constant k, so the formula becomes C = 8.85418 k A / d picofarads, if the area A

    is in square meters and the distance d is in meters.

    For the second part: the quantity of charge Q equals the product of the capacitance

    and the applied voltage; that is: Q = CV, where Q is the charge in coulombs, C the

    capacitance in farads, and V the applied voltage (in volts).

    EDIT: The capacitance is independent of the applied voltage; it just depends on the

    physical dimensions of the capacitor and the insulating material used.

    A higher voltage will put a higher charge on it, but its capacitance stays the same.

    Source(s): college physics - the definition of a farad is the amount of capacitance to which one volt will put a charge of one coulomb
  • 9 years ago

    Capacitance in a parallel plate capacitor is:

    dielectric constant * epsilon nought * area/ distance

    Its a geometry thing and a whats in between the plates thing.

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