Spherical Capacitors

8/3/2019 Spherical Capacitors

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Spherical Capacitors

Consider an isolated, initially uncharged, metal conductor. After the first small amount of

charge, q, is placed on the conductor, its voltage becomes as compared to V = 0 at

infinity. To further charge the conductor, work must be done to bring increments of charge,dq, to its surface:

The amount of work required to bring in each additional charged-increment, dq, increases as the

spherical conductor becomes more highly charged. The total electric potential energy of the

conductor can be calculated by

The capacitance of the spherical conductor can be calculated as

Notice that a spherical conductor's capacitance is totally dependent on the sphere's radius.


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Refer to the following information for the next three questions.

A spherical conductor has a diameter of 10 cm.

What is its capacitance in farads?

5.56 x 10-12

F

If the conductor holds 6 C of charge, then what is the electric potential at its surface?

1.08 x 106

V

How much work was required to charge the capacitor?

3.24 J

Parallel Plate Capacitors

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.


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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 runparallel 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 enclosed in our gaussian box equals

Thus,

We also know that the potential difference across the plates is equal to

Since plate B is defined to be at V = 0,


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we can rewrite this as

Substituting into Q = CV yields

To relate the energy per unit volume stored in a capacitor to the magnitude of its electric field,

we will build on our relationship for the energy stored in a capacitor developed in a previouslesson.
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Cylindrical Capacitors

Now let's consider the geometry of a cylindrical capacitor. Suppose that our capacitor is

composed of an inner cylinder with radius a enclosed by an outer cylinder with radius b.


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

Again, we will use Gauss' Law to evaluate the electric field between the plates by using agaussian surface that is cylindrical in shape and of length L. The cylinder has a uniform charge

per unit length of .


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We also know that the potential difference across the cylinders is equal to

Since the outer plate is negative, its voltage can be set equal to 0, and we can state that the

potential difference across the capacitors equals

Returning to Q = CV


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This represents the capacitance per unit length of our cylindrical capacitor. An excellent example

of a cylindrical capacitor is the coaxial cable used in cable TV system

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Spherical Capacitors