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APC Resource Lesson
Capacitors: Parallel Plates and Cylinders


In this lesson we will derive the equations for capacitance based on two special types of geometries: capacitors with parallel plates and those with cylindrical cables.

Let us start with parallel plates.

Since we know that the basic relationship Q = CV, we must obtain expressions for Q and V to evaluate C.

Using Gauss' Law,

We can evaluate E, the electric field between the plates once we employ an appropriate gaussian surface. In this case, we will use a box with one side embedded within the top plate.

This box has six faces: a top, a bottom, left side, right side, front surface and back surface. Since the top surface is embedded within the metal plate, no field lines will pass through it since under electrostatic conditions there are no field lines within a conductor. Field lines will only run parallel to the area vector of the bottom surface. They will be perpendicular to the area vectors of the other four sides. Thus,

The total charge on each plate equals

Thus,

If a positive test charge were to be moved between the plates, from A to B, its electric potential energy (EPE) would decrease while its kinetic energy (KE) would increase.

Substituting into Q = CV yields

Now let's consider the geometry of a cylindrical capacitor. Suppose that our capacitor is composed of an inner cylinder (plate) with radius a enclosed by an outer cylinder (plate) with radius b.

Since we know that the basic relationship Q = CV, we must obtain expressions for Q and V to evaluate C.

We will use Gauss' Law to evaluate the electric field between the plates by using a gaussian surface that is cylindrical in shape of length L.

Since our cylinders have a uniform charge distribution,

If a positive test charge were to be moved between the plates, from A to B, its electric potential energy (EPE) would decrease while its kinetic energy (KE) would increase.

Multiplying through by negative 1 yields:

Returning to Q = CV

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