# Thread: Fsc 2nd Year chapter No :1 | ELECTROSTATICS

1. ## Fsc 2nd Year chapter No :1 | ELECTROSTATICS

Fsc 2nd Year chapter No :1 | ELECTROSTATICS

Topic No:14

CAPACITANCE OF A PARALLEL PLATE CAPACITOR

Parallel Plate Capacitor

The capacitance of flat, parallel metallic plates of area A and separation d is given by the expression above where:
= permittivity of space and k = relative permittivity of the dielectric material between the plates.
k=1 for free space, k>1 for all media, approximately =1 for air.
The Farad, F, is the SI unit for capacitance, and from the definition of capacitance is seen to be equal to a Coulomb/Volt.
Any of the active parameters in the expression below can be calculated by clicking on it. Default values will be provided for any parameters left unspecified, but all parameters can be changed. After editing data, you must click on the desired parameter to calculate; values will not automatically be forced to be consistent.

Capacitance of Parallel Plates

The electric field between two large parallel plates is given by

The voltage difference between the two plates can be expressed in terms of the work done on a positive test charge q when it moves from the positive to the negative plate.
It then follows from the definition of capacitance that
Electric Field: Parallel Plates

If oppositely charges parallel conducting plates are treated like infinite planes (neglecting fringing), then Gauss' law can be used to calculate the electric field between the plates. Presuming the plates to be at equilibrium with zero electric field inside the conductors, then the result from a charged conducting surface can be used:
This is also consistent with treating the charge layers as two charge sheets with electric field

in both directions.

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